2.3 The Limit Laws
Learning Objectives
- Recognize the basic limit laws.
- Use the limit laws to evaluate the limit of a function.
- Evaluate the limit of a function by factoring.
- Use the limit laws to evaluate the limit of a polynomial or rational function.
- Evaluate the limit of a function by factoring or by using conjugates.
- Evaluate the limit of a function by using the squeeze theorem.
In the previous section we estimated limits the slow way — by tabulating values or squinting at graphs. That is fine for building intuition, but it is no way to do real work. In this section we trade pictures for algebra. We collect a small set of limit laws — rules that say "the limit of a sum is the sum of the limits," and so on — and then we drive limits exactly by manipulating expressions instead of guessing from tables. By the end of the section you will be evaluating limits the same way you evaluate ordinary algebraic expressions, with the limit symbol just along for the ride. A Student Project at the end uses these new tools to chase down the area of a circle, the way Archimedes essentially did two thousand years ago — by approximating the circle with regular polygons and letting the vertex angle shrink toward zero.
We begin by restating two limits from the previous section that we will treat as basic building blocks. These two facts plus the limit laws are the entire foundation for the rest of the section.
2.3.1 Evaluating Limits with the Limit Laws
The first two limits below appeared in the previous section. They look almost trivial, but they are the seeds from which every other algebraic limit grows.
For any real number \(a\) and any constant \(c\),
$$ \lim_{x\to a}x=a \qquad\text{and}\qquad \lim_{x\to a}c=c. $$Every limit law that follows assumes you already know the limits of the two simplest functions in the world: the identity function \(f(x)=x\) and a constant function \(f(x)=c\). If those two are nailed down, the limit laws let you bootstrap up to any polynomial, any rational function, any algebraic combination — without ever building another table.
The first says: as \(x\) marches toward \(a\), the identity function \(f(x)=x\) marches right with it. The second says: a constant function never moves, so its limit is just the constant. Both are obvious once you remember that a limit asks about the value the function settles on, and these two functions are the easiest possible cases.
Use the limit laws to evaluate \(\displaystyle\lim_{x\to 6}(2x-1)\sqrt{x+4}.\) In each step, indicate the limit law applied.
Solution
Step 1 — Apply the product law. Both factors will have limits at \(x=6\), so we can split the product:
$$ \lim_{x\to 6}(2x-1)\sqrt{x+4}=\Bigl(\lim_{x\to 6}(2x-1)\Bigr)\cdot \Bigl(\lim_{x\to 6}\sqrt{x+4}\Bigr). $$Step 2 — First factor: sum, constant multiple, basic limits.
$$ \lim_{x\to 6}(2x-1)=2\cdot \lim_{x\to 6}x-\lim_{x\to 6}1=2(6)-1=11. $$Step 3 — Second factor: root law, then sum and basic limits. The radicand limit is \(6+4=10>0\), so the root law applies (index 2 is even):
$$ \lim_{x\to 6}\sqrt{x+4}=\sqrt{\lim_{x\to 6}(x+4)}=\sqrt{6+4}=\sqrt{10}. $$Step 4 — Recombine via the product law.
$$ \lim_{x\to 6}(2x-1)\sqrt{x+4}=11\sqrt{10}. $$Answer: \(11\sqrt{10}.\)
Evaluate each of the following limits using the Basic Limit Results.
- 1. \(\displaystyle\lim_{x\to 2}x\)
- 2. \(\displaystyle\lim_{x\to 2}5\)
Solution
Step 1 — Identify the basic form. The first limit is the identity function evaluated at \(a=2\). The second is a constant function whose value is always 5.
Step 2 — Apply Theorem 2.3.1.
- 1. \(\displaystyle\lim_{x\to 2}x=2\) — the limit of \(x\) as \(x\to a\) is \(a\).
- 2. \(\displaystyle\lim_{x\to 2}5=5\) — the limit of a constant is the constant itself.
Answer: \(\;\displaystyle\lim_{x\to 2}x=2\) and \(\displaystyle\lim_{x\to 2}5=5\).
Now we take a look at the limit laws — the rules that let us break a complicated limit into pieces, take the limit of each piece, and stitch the answers back together. Proofs of these laws use the \(\varepsilon\)–\(\delta\) definition and are deferred to later in the chapter; for now we will take them as the operational rulebook for limits.
Let \(f(x)\) and \(g(x)\) be defined for all \(x\ne a\) over some open interval containing \(a\). Assume \(L\) and \(M\) are real numbers such that
$$ \lim_{x\to a}f(x)=L \qquad\text{and}\qquad \lim_{x\to a}g(x)=M. $$Let \(c\) be a constant. Then each of the following holds:
- Sum law: \(\displaystyle\lim_{x\to a}\bigl(f(x)+g(x)\bigr)=L+M.\)
- Difference law: \(\displaystyle\lim_{x\to a}\bigl(f(x)-g(x)\bigr)=L-M.\)
- Constant multiple law: \(\displaystyle\lim_{x\to a}cf(x)=cL.\)
- Product law: \(\displaystyle\lim_{x\to a}\bigl(f(x)\cdot g(x)\bigr)=L\cdot M.\)
- Quotient law: \(\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{L}{M}\) for \(M\ne 0\).
- Power law: \(\displaystyle\lim_{x\to a}\bigl(f(x)\bigr)^n=L^n\) for every positive integer \(n\).
- Root law: \(\displaystyle\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{L}\) for all \(L\) when \(n\) is odd, and for \(L\ge 0\) (with \(f(x)\ge 0\)) when \(n\) is even.
Once you know the individual limits exist, you can move \(\lim_{x\to a}\) inside sums, differences, products, quotients (denominator nonzero), powers, and \(n\)th roots — exactly the algebraic moves you already use on numbers. The catch is the "once you know they exist" clause: every law assumes the pieces have limits, and every step needs to be checked before you take it.
We will now practice applying these laws to evaluate a limit step by step.
Use the limit laws to evaluate \(\displaystyle\lim_{x\to -3}(4x+2).\)
Solution
Apply the laws one step at a time. At each step we have to check that the smaller limits actually exist before we are allowed to split them apart.
Step 1 — Apply the sum law.
$$ \lim_{x\to -3}(4x+2)=\lim_{x\to -3}4x+\lim_{x\to -3}2. $$Step 2 — Apply the constant multiple law to the first piece.
$$ =4\cdot \lim_{x\to -3}x+\lim_{x\to -3}2. $$Step 3 — Apply the basic limit results and simplify.
$$ =4\cdot(-3)+2=-10. $$Answer: \(\displaystyle\lim_{x\to -3}(4x+2)=-10.\)
Use the limit laws to evaluate \(\displaystyle\lim_{x\to 2}\dfrac{2x^2-3x+1}{x^3+4}.\)
Solution
Several laws have to fire here. As we rewrite the limit in terms of smaller limits, each new limit has to exist (and any denominator we form has to be nonzero) for the next law to apply.
Step 1 — Apply the quotient law. First check the denominator: at \(x=2\), \(x^3+4=8+4=12\ne 0\), so the denominator-limit is nonzero and we are allowed to split:
$$ \lim_{x\to 2}\frac{2x^2-3x+1}{x^3+4}=\frac{\displaystyle\lim_{x\to 2}(2x^2-3x+1)}{\displaystyle\lim_{x\to 2}(x^3+4)}. $$Step 2 — Apply the sum and constant multiple laws.
$$ =\frac{2\cdot \lim_{x\to 2}x^2-3\cdot \lim_{x\to 2}x+\lim_{x\to 2}1}{\lim_{x\to 2}x^3+\lim_{x\to 2}4}. $$Step 3 — Apply the power law.
$$ =\frac{2\bigl(\lim_{x\to 2}x\bigr)^2-3\cdot \lim_{x\to 2}x+\lim_{x\to 2}1}{\bigl(\lim_{x\to 2}x\bigr)^3+\lim_{x\to 2}4}. $$Step 4 — Apply the basic limit results and simplify.
$$ =\frac{2(4)-3(2)+1}{(2)^3+4}=\frac{3}{12}=\frac{1}{4}. $$Answer: \(\displaystyle\lim_{x\to 2}\dfrac{2x^2-3x+1}{x^3+4}=\dfrac{1}{4}.\)
2.3.2 Limits of Polynomial and Rational Functions
If you stare at the previous examples you start to notice a pattern: in every case the answer came out to be the function value at \(x=a\). That is not an accident for polynomials and rational functions — it is a theorem.
Let \(p(x)\) and \(q(x)\) be polynomial functions, and let \(a\) be a real number. Then
$$ \lim_{x\to a}p(x)=p(a), $$and, provided \(q(a)\ne 0\),
$$ \lim_{x\to a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}. $$Polynomial functions are continuous everywhere and rational functions are continuous everywhere their denominator is nonzero. For these functions, \(\displaystyle\lim_{x\to a}f(x)=f(a)\) — you just substitute. This is the most useful rule of thumb in the section. The interesting limits are the ones where you cannot just plug in, because the function has a hole or a blow-up at \(a\); those need the techniques in the next subsection.
To see why the polynomial part of this theorem is true, take a general polynomial \(p(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0.\) Apply the sum law, then the constant multiple law, then the power law, and finally the basic limit results:
$$ \begin{aligned} \lim_{x\to a}p(x)&=\lim_{x\to a}\bigl(c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0\bigr)\\ &=c_n\bigl(\lim_{x\to a}x\bigr)^n+c_{n-1}\bigl(\lim_{x\to a}x\bigr)^{n-1}+\cdots+c_1\lim_{x\to a}x+\lim_{x\to a}c_0\\ &=c_na^n+c_{n-1}a^{n-1}+\cdots+c_1a+c_0\\ &=p(a). \end{aligned} $$The rational case then follows from the quotient law: if \(q(a)\ne 0\), the denominator-limit is nonzero and we can split.
$$ \lim_{x\to a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}. $$Evaluate \(\displaystyle\lim_{x\to -2}\bigl(3x^3-2x+7\bigr).\)
Solution
Step 1 — Recognize a polynomial. \(p(x)=3x^3-2x+7\) is a polynomial, so by Theorem 2.3.3 we just plug in.
Step 2 — Substitute \(x=-2\).
$$ p(-2)=3(-2)^3-2(-2)+7=3(-8)+4+7=-24+11=-13. $$Answer: \(\displaystyle\lim_{x\to -2}\bigl(3x^3-2x+7\bigr)=-13.\)
Evaluate \(\displaystyle\lim_{x\to 3}\dfrac{2x^2-3x+1}{5x+4}.\)
Solution
Step 1 — Check the denominator at \(x=3\). \(5(3)+4=19\ne 0\), so 3 is in the domain of the rational function. Theorem 2.3.3 applies.
Step 2 — Substitute.
$$ \lim_{x\to 3}\frac{2x^2-3x+1}{5x+4}=\frac{2(3)^2-3(3)+1}{5(3)+4}=\frac{18-9+1}{19}=\frac{10}{19}. $$Answer: \(\displaystyle\lim_{x\to 3}\dfrac{2x^2-3x+1}{5x+4}=\dfrac{10}{19}.\)
2.3.3 Additional Limit Evaluation Techniques
Substituting works whenever \(a\) is in the domain. The interesting limits — and the ones that drive every derivative — are the ones where \(f(a)\) is undefined but the limit still exists. The key observation is small but powerful:
If two functions agree everywhere except at \(x=a\), they have the same limit at \(a\). The single missing point is invisible to the limit. So our strategy for indeterminate-form limits is: rewrite the expression so the troublesome factor cancels, then take the limit of the simpler expression.
Stated more carefully: if \(f(x)=g(x)\) for all \(x\ne a\) over some open interval containing \(a\), then \(\displaystyle\lim_{x\to a}f(x)=\lim_{x\to a}g(x).\)
To see this idea in action, consider \(\displaystyle\lim_{x\to 1}\dfrac{x^2-1}{x-1}.\) The function
$$ f(x)=\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1} $$is undefined at \(x=1\) (division by zero), while \(g(x)=x+1\) is defined everywhere. But \(f(x)=g(x)\) for every other \(x\). Their graphs agree except for the single hole at \(x=1\), as shown in Figure 2.24.
Figure 2.24 — The graphs of \(f(x)\) and \(g(x)\) are identical for all \(x\ne 1.\) Their limits at 1 are equal.
So
$$ \lim_{x\to 1}\frac{x^2-1}{x-1}=\lim_{x\to 1}\frac{(x-1)(x+1)}{x-1}=\lim_{x\to 1}(x+1)=2. $$This limit has the form \(\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}\) where both \(f(x)\to 0\) and \(g(x)\to 0\). We say \(f(x)/g(x)\) has the indeterminate form \(0/0\) at \(a\) — "indeterminate" because the answer is not determined by the symbols alone (a \(0/0\) form can resolve to any finite number, to \(\pm\infty\), or to DNE, depending on the function). The next strategy box collects the standard moves.
Problem-Solving Strategy — Calculating a Limit When \(f(x)/g(x)\) has the Indeterminate Form \(0/0\)
1. First, check whether direct substitution applies. If \(f(a)\) and \(g(a)\) are both nonzero (or the function is otherwise continuous at \(a\)), just plug in via Theorem 2.3.3.
2. If substitution gives \(0/0\), do not give up — rewrite. Try one of the following moves, in roughly this order of preference:
- Factor and cancel when both numerator and denominator are polynomials (or factor algebraically).
- Multiply by a conjugate when a square root is involved — multiply numerator and denominator by the conjugate of the radical expression.
- Simplify a complex fraction — clear inner denominators by multiplying numerator and denominator by their least common denominator.
3. After cancelling the troublesome factor, retake the limit on the simplified expression using Theorem 2.3.3 (direct substitution).
The next three examples walk through these three moves one at a time. Example 2.3.5 illustrates the factor-and-cancel technique; Example 2.3.6 shows multiplying by a conjugate; Example 2.3.7 simplifies a complex fraction.
Evaluate \(\displaystyle\lim_{x\to -3}\dfrac{x^2+4x+3}{x^2-9}.\)
Solution
Step 1 — Detect the indeterminate form. At \(x=-3\): numerator \(=9-12+3=0\); denominator \(=9-9=0\). Form is \(0/0\) — factor and cancel.
Step 2 — Factor.
$$ \frac{x^2+4x+3}{x^2-9}=\frac{(x+1)(x+3)}{(x-3)(x+3)}. $$Step 3 — Cancel \((x+3)\) and substitute. For \(x\ne -3\), the expression equals \(\dfrac{x+1}{x-3}\), so
$$ \lim_{x\to -3}\frac{x^2+4x+3}{x^2-9}=\lim_{x\to -3}\frac{x+1}{x-3}=\frac{-3+1}{-3-3}=\frac{-2}{-6}=\frac{1}{3}. $$Answer: \(\dfrac{1}{3}.\)
Evaluate \(\displaystyle\lim_{x\to 3}\dfrac{x^2-3x}{2x^2-5x-3}.\)
Solution
Step 1 — Try direct substitution; detect the indeterminate form. Plugging in \(x=3\) gives \(\dfrac{9-9}{18-15-3}=\dfrac{0}{0}\). Substitution fails; we need to rewrite. Both top and bottom are polynomials, so factoring is the natural first move:
$$ \lim_{x\to 3}\frac{x^2-3x}{2x^2-5x-3}=\lim_{x\to 3}\frac{x(x-3)}{(x-3)(2x+1)}. $$Step 2 — Cancel the troublesome factor. For all \(x\ne 3\), \(\dfrac{x(x-3)}{(x-3)(2x+1)}=\dfrac{x}{2x+1}\). The limit only cares about \(x\ne 3\), so:
$$ \lim_{x\to 3}\frac{x(x-3)}{(x-3)(2x+1)}=\lim_{x\to 3}\frac{x}{2x+1}. $$Step 3 — Take the limit of the simplified expression. \(\dfrac{x}{2x+1}\) is a rational function with \(2(3)+1=7\ne 0\) at \(x=3\), so we can just substitute:
$$ \lim_{x\to 3}\frac{x}{2x+1}=\frac{3}{2(3)+1}=\frac{3}{7}. $$Answer: \(\displaystyle\lim_{x\to 3}\dfrac{x^2-3x}{2x^2-5x-3}=\dfrac{3}{7}.\)
Evaluate \(\displaystyle\lim_{x\to 5}\dfrac{\sqrt{x-1}-2}{x-5}.\)
Solution
Step 1 — Detect the indeterminate form. At \(x=5\): numerator \(=\sqrt{4}-2=0\); denominator \(=0\). Multiply by the conjugate \(\sqrt{x-1}+2\):
$$ \frac{\sqrt{x-1}-2}{x-5}\cdot \frac{\sqrt{x-1}+2}{\sqrt{x-1}+2}=\frac{(x-1)-4}{(x-5)(\sqrt{x-1}+2)}=\frac{x-5}{(x-5)(\sqrt{x-1}+2)}. $$Step 2 — Cancel \((x-5)\) and substitute.
$$ \lim_{x\to 5}\frac{1}{\sqrt{x-1}+2}=\frac{1}{\sqrt{4}+2}=\frac{1}{4}. $$Answer: \(\dfrac{1}{4}.\)
Evaluate \(\displaystyle\lim_{x\to -1}\dfrac{\sqrt{x+2}-1}{x+1}.\)
Solution
Step 1 — Detect the indeterminate form. At \(x=-1\): numerator \(=\sqrt{1}-1=0\); denominator \(=0\). Form is \(0/0\). A square root in the numerator is a strong hint to multiply by the conjugate \(\sqrt{x+2}+1\):
$$ \lim_{x\to -1}\frac{\sqrt{x+2}-1}{x+1}=\lim_{x\to -1}\frac{\sqrt{x+2}-1}{x+1}\cdot \frac{\sqrt{x+2}+1}{\sqrt{x+2}+1}. $$Step 2 — Multiply out the numerator only. \((\sqrt{x+2}-1)(\sqrt{x+2}+1)=(x+2)-1=x+1\). Leave the denominator factored — we want the \((x+1)\) to cancel:
$$ =\lim_{x\to -1}\frac{x+1}{(x+1)(\sqrt{x+2}+1)}. $$Step 3 — Cancel \((x+1)\). For \(x\ne -1\),
$$ =\lim_{x\to -1}\frac{1}{\sqrt{x+2}+1}. $$Step 4 — Apply the limit laws. The radicand limit \(\to 1>0\), so the root law applies and we can just substitute:
$$ \lim_{x\to -1}\frac{1}{\sqrt{x+2}+1}=\frac{1}{\sqrt{1}+1}=\frac{1}{2}. $$Answer: \(\displaystyle\lim_{x\to -1}\dfrac{\sqrt{x+2}-1}{x+1}=\dfrac{1}{2}.\)
Evaluate \(\displaystyle\lim_{x\to -3}\dfrac{\frac{1}{x+2}+1}{x+3}.\)
Solution
Step 1 — Detect the indeterminate form. At \(x=-3\): numerator \(=\tfrac{1}{-1}+1=0\); denominator \(=0\).
Step 2 — Clear the inner denominator by multiplying by \((x+2)/(x+2)\).
$$ \frac{\tfrac{1}{x+2}+1}{x+3}\cdot \frac{x+2}{x+2}=\frac{1+(x+2)}{(x+3)(x+2)}=\frac{x+3}{(x+3)(x+2)}. $$Step 3 — Cancel \((x+3)\) and substitute.
$$ \lim_{x\to -3}\frac{1}{x+2}=\frac{1}{-3+2}=-1. $$Answer: \(-1.\)
The next example does not fit cleanly into any of the three patterns above — neither piece even has a limit at the target point. But with a little algebra, we can recombine the two pieces into a single fraction and then return to the techniques we already have.
Evaluate \(\displaystyle\lim_{x\to 1}\dfrac{\frac{1}{x+1}-\frac{1}{2}}{x-1}.\)
Solution
Step 1 — Detect the indeterminate form. At \(x=1\): numerator \(=\tfrac12-\tfrac12=0\); denominator \(=0\). The expression is a complex fraction; clear the inner denominators by multiplying numerator and denominator by \(2(x+1)\):
$$ \lim_{x\to 1}\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}=\lim_{x\to 1}\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}\cdot \frac{2(x+1)}{2(x+1)}. $$Step 2 — Multiply through the numerator only. Distribute \(2(x+1)\) over \(\tfrac{1}{x+1}-\tfrac{1}{2}\): we get \(2-(x+1)\). Leave the denominator factored:
$$ =\lim_{x\to 1}\frac{2-(x+1)}{2(x-1)(x+1)}. $$Step 3 — Simplify the numerator.
$$ =\lim_{x\to 1}\frac{-x+1}{2(x-1)(x+1)}. $$Step 4 — Factor out \(-1\) so the troublesome factor matches.
$$ =\lim_{x\to 1}\frac{-(x-1)}{2(x-1)(x+1)}. $$Step 5 — Cancel \((x-1)\). For \(x\ne 1\),
$$ =\lim_{x\to 1}\frac{-1}{2(x+1)}. $$Step 6 — Substitute.
$$ \lim_{x\to 1}\frac{-1}{2(x+1)}=\frac{-1}{2(2)}=-\frac{1}{4}. $$Answer: \(\displaystyle\lim_{x\to 1}\dfrac{\frac{1}{x+1}-\frac{1}{2}}{x-1}=-\dfrac{1}{4}.\)
Evaluate \(\displaystyle\lim_{x\to 3}\!\left(\dfrac{1}{x-3}-\dfrac{4}{x^2-2x-3}\right).\)
Solution
Step 1 — Combine into a single fraction. Factor the second denominator: \(x^2-2x-3=(x-3)(x+1)\). Common denominator \((x-3)(x+1)\):
$$ \frac{1}{x-3}-\frac{4}{(x-3)(x+1)}=\frac{(x+1)-4}{(x-3)(x+1)}=\frac{x-3}{(x-3)(x+1)}. $$Step 2 — Cancel \((x-3)\) and substitute.
$$ \lim_{x\to 3}\frac{1}{x+1}=\frac{1}{4}. $$Answer: \(\dfrac{1}{4}.\)
Now let us revisit one-sided limits. The limit laws apply to one-sided limits with one small modification: we only need the function defined on an open interval to the appropriate side of \(a\). For \(\displaystyle\lim_{x\to a^-}h(x)\) we need \(h(x)\) defined on some open interval \((b,a)\); for \(\displaystyle\lim_{x\to a^+}h(x)\) we need \(h(x)\) defined on some \((a,c)\). The next example shows this matters.
Evaluate \(\displaystyle\lim_{x\to 0}\!\left(\dfrac{1}{x}+\dfrac{5}{x(x-5)}\right).\)
Solution
Step 1 — Why the sum law fails. Both \(1/x\) and \(5/(x(x-5))\) blow up as \(x\to 0\) (their limits do not exist). The sum law requires both pieces to have real-number limits, so it does not apply. We have to combine the fractions first.
Step 2 — Combine. Common denominator \(x(x-5)\):
$$ \frac{1}{x}+\frac{5}{x(x-5)}=\frac{(x-5)+5}{x(x-5)}=\frac{x}{x(x-5)}. $$Step 3 — Cancel \(x\). For \(x\ne 0\),
$$ \frac{x}{x(x-5)}=\frac{1}{x-5}. $$Step 4 — Take the limit of the simplified expression. \(1/(x-5)\) is continuous at \(x=0\):
$$ \lim_{x\to 0}\left(\frac{1}{x}+\frac{5}{x(x-5)}\right)=\lim_{x\to 0}\frac{1}{x-5}=-\frac{1}{5}. $$Answer: \(-\dfrac{1}{5}.\)
Graph \(f(x)=\begin{cases}-x-2 & \text{if } x<-1 \\ 2 & \text{if } x=-1 \\ x^3 & \text{if } x>-1\end{cases}\) and evaluate \(\displaystyle\lim_{x\to -1^-}f(x).\)
Solution
Step 1 — Identify the relevant branch. For \(x<-1\), \(f(x)=-x-2\) (a line of slope \(-1\)). For \(x>-1\), \(f(x)=x^3\). The value \(f(-1)=2\) is an isolated point — it does not affect the limit.
Step 2 — Evaluate the one-sided limit. Use the linear branch:
$$ \lim_{x\to -1^-}f(x)=\lim_{x\to -1^-}(-x-2)=-(-1)-2=-1. $$Step 3 — Sketch. Draw \(y=-x-2\) for \(x<-1\) ending in an open circle at \((-1,-1)\); plot the isolated dot at \((-1,2)\); draw \(y=x^3\) for \(x>-1\) starting at an open circle at \((-1,-1)\).
Answer: \(-1.\)
Finally, we look at the form \(K/0\) with \(K\ne 0\). When the numerator approaches a nonzero number and the denominator approaches 0, the quotient law breaks (you cannot divide by 0). The magnitude of the quotient becomes infinite, so the limit is \(+\infty\), \(-\infty\), or does not exist — and we have to use the sign of each piece to figure out which.
Evaluate each of the following limits, if possible.
- 1. \(\displaystyle\lim_{x\to 3^-}\sqrt{x-3}\)
- 2. \(\displaystyle\lim_{x\to 3^+}\sqrt{x-3}\)
Figure 2.25 shows the function \(f(x)=\sqrt{x-3}.\)
Figure 2.25 — The graph shows the function \(f(x)=\sqrt{x-3}.\)
Solution
Step 1 — Identify the domain. \(f(x)=\sqrt{x-3}\) is defined on \([3,+\infty)\). Nothing exists to the left of 3.
Step 2 — Left-hand limit. Since \(f\) is not defined on any open interval \((b,3)\) to the left of 3, the limit laws do not apply, and in fact \(\displaystyle\lim_{x\to 3^-}\sqrt{x-3}\) does not exist.
Step 3 — Right-hand limit. \(f\) is defined on \((3,c)\) for any \(c>3\). The root law applies (radicand limit \(\to 0\), with \(f(x)\ge 0\)):
$$ \lim_{x\to 3^+}\sqrt{x-3}=\sqrt{\lim_{x\to 3^+}(x-3)}=\sqrt{0}=0. $$Answer: Part 1 does not exist; part 2 equals \(0\).
The next example uses one-sided limits to decide whether the two-sided limit of a piecewise function exists.
Evaluate \(\displaystyle\lim_{x\to 1}\dfrac{x+2}{(x-1)^2}.\)
Solution
Step 1 — Read the form. At \(x=1\): numerator \(=3\); denominator \(=0\). Form is \(3/0\) — magnitude explodes.
Step 2 — Determine the sign of the denominator. \((x-1)^2\) is a square, so it is positive for every \(x\ne 1\). The denominator approaches 0 from above on both sides.
Step 3 — Combine signs. Positive numerator \(\to 3\), positive denominator \(\to 0^+\):
$$ \lim_{x\to 1}\frac{x+2}{(x-1)^2}=+\infty. $$Answer: \(+\infty.\)
For \(f(x)=\begin{cases}4x-3 & \text{if } x<2 \\ (x-3)^2 & \text{if } x\ge 2\end{cases},\) evaluate
- 1. \(\displaystyle\lim_{x\to 2^-}f(x)\)
- 2. \(\displaystyle\lim_{x\to 2^+}f(x)\)
- 3. \(\displaystyle\lim_{x\to 2}f(x)\)
Figure 2.26 shows \(f(x).\)
Figure 2.26 — This graph shows a function \(f(x).\)
Solution
Step 1 — Left-hand limit. For \(x\) in \((-\infty,2)\), \(f(x)=4x-3\). Replace \(f\) by this branch and apply the limit laws:
$$ \lim_{x\to 2^-}f(x)=\lim_{x\to 2^-}(4x-3)=4(2)-3=5. $$Step 2 — Right-hand limit. For \(x\) in \([2,+\infty)\), \(f(x)=(x-3)^2\). Replace \(f\) by this branch:
$$ \lim_{x\to 2^+}f(x)=\lim_{x\to 2^+}(x-3)^2=(2-3)^2=1. $$Step 3 — Two-sided limit. Since the left- and right-hand limits disagree (\(5\ne 1\)), the two-sided limit \(\displaystyle\lim_{x\to 2}f(x)\) does not exist.
Answer: \(5,\ 1,\) and DNE.
Evaluate \(\displaystyle\lim_{x\to 2^-}\dfrac{x-3}{x^2-2x}.\)
Solution
Step 1 — Plug in and read the form. At \(x=2\): numerator \(=-1\); denominator \(=4-4=0\). Form is \(-1/0\); the magnitude of the quotient explodes. To pin down the sign we factor the denominator:
$$ \lim_{x\to 2^-}\frac{x-3}{x^2-2x}=\lim_{x\to 2^-}\frac{x-3}{x(x-2)}. $$Step 2 — Peel off the troublesome factor. Write the expression as a product so that one factor has a finite limit and the other has the blow-up:
$$ =\lim_{x\to 2^-}\frac{x-3}{x}\cdot \frac{1}{x-2}. $$Step 3 — Take each factor's limit. \(\displaystyle\lim_{x\to 2^-}\dfrac{x-3}{x}=\dfrac{2-3}{2}=-\dfrac{1}{2}.\) For \(\dfrac{1}{x-2}\) as \(x\to 2^-\), the denominator \(x-2\) is a small negative number, so \(\dfrac{1}{x-2}\to -\infty.\)
Step 4 — Multiply the signs and magnitudes. Negative finite \(\times\) negative infinity = positive infinity:
$$ \lim_{x\to 2^-}\frac{x-3}{x^2-2x}=+\infty. $$Answer: \(+\infty.\)
2.3.4 The Squeeze Theorem
The techniques above work cleanly on algebraic functions, but they cannot evaluate even basic trig limits like \(\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}.\) The next theorem — the squeeze theorem — is the right tool for these. The idea: if you can trap an unknown function \(g(x)\) between two known functions \(f(x)\) and \(h(x)\) that agree on a common limit \(L\) at \(a\), then \(g(x)\) has no choice but to share that limit too. Figure 2.27 shows the picture.
Figure 2.27 — The Squeeze Theorem applies when \(f(x)\le g(x)\le h(x)\) and \(\lim_{x\to a}f(x)=\lim_{x\to a}h(x).\)
Let \(f(x),\;g(x),\) and \(h(x)\) be defined for all \(x\ne a\) over an open interval containing \(a\). If
$$ f(x)\le g(x)\le h(x) \quad\text{for all } x\ne a \text{ in an open interval containing } a, $$Picture two functions \(f\) and \(h\) that pinch together at \(x=a\), heading to the same height \(L\). Any function \(g\) caught between them on a neighborhood of \(a\) is squeezed into the same point — it cannot wiggle to a different value without crossing one of the trapping functions. The theorem turns this intuitive picture into a rigorous tool.
and
$$ \lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L, $$where \(L\) is a real number, then
$$ \lim_{x\to a}g(x)=L. $$Use the squeeze theorem to evaluate \(\displaystyle\lim_{x\to 0}x^2\sin\!\frac{1}{x}.\)
Solution
Step 1 — Trap. Because \(-1\le \sin(1/x)\le 1\) for every \(x\ne 0\), multiplying by \(x^2\ge 0\) preserves the inequality:
$$ -x^2\le x^2\sin\!\frac{1}{x}\le x^2. $$Step 2 — Outer limits agree. \(\displaystyle\lim_{x\to 0}(-x^2)=0=\lim_{x\to 0}x^2.\)
Step 3 — Squeeze.
$$ \lim_{x\to 0}x^2\sin\!\frac{1}{x}=0. $$Answer: \(0.\)
We now use the squeeze theorem to tackle several very important trigonometric limits. These will feel like a long detour, but the limits we are about to nail down — \(\displaystyle\lim_{\theta\to 0}\sin\theta\), \(\displaystyle\lim_{\theta\to 0}\cos\theta\), and especially \(\displaystyle\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta}\) — are the seeds of every derivative and integral of a trigonometric function later in the book.
Start with \(\displaystyle\lim_{\theta\to 0}\sin\theta.\) Look at the unit circle in Figure 2.29. The value \(\sin\theta\) is the \(y\)-coordinate of the point on the unit circle at angle \(\theta\); it is the length of the blue line segment. The radian measure of \(\theta\) is the arc length on the unit circle, drawn alongside. From the picture, for \(0<\theta<\dfrac{\pi}{2}\) we have \(0<\sin\theta<\theta.\)
Figure 2.29 — The sine function is shown as a line on the unit circle.
Both outer functions head to 0:
$$ \lim_{\theta\to 0^+}0=0 \qquad\text{and}\qquad \lim_{\theta\to 0^+}\theta=0, $$so by the squeeze theorem,
$$ \lim_{\theta\to 0^+}\sin\theta=0. $$To see that \(\displaystyle\lim_{\theta\to 0^-}\sin\theta=0\) as well, take \(-\tfrac{\pi}{2}<\theta<0\). Then \(0<-\theta<\tfrac{\pi}{2}\), so \(0<\sin(-\theta)<-\theta\). But \(\sin(-\theta)=-\sin\theta\), so \(0<-\sin\theta<-\theta\), i.e., \(0>\sin\theta>\theta.\) Another squeeze gives the left-side limit. Combining the two sides,
$$ \lim_{\theta\to 0}\sin\theta=0. $$Next, the identity \(\cos\theta=\sqrt{1-\sin^2\theta}\) on \(-\tfrac{\pi}{2}<\theta<\tfrac{\pi}{2}\) gives
$$ \lim_{\theta\to 0}\cos\theta=\lim_{\theta\to 0}\sqrt{1-\sin^2\theta}=\sqrt{1-0^2}=1. $$Now to the headline limit: \(\displaystyle\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta}.\) We use the unit circle in Figure 2.30. The figure adds a tangent triangle to the previous picture: the side opposite \(\theta\) in this new triangle has length \(\tan\theta\). From the picture, for \(0<\theta<\tfrac{\pi}{2}\) we have \(\sin\theta<\theta<\tan\theta.\)
Figure 2.30 — The sine and tangent functions are shown as lines on the unit circle.
Divide every part of the inequality by \(\sin\theta\) (positive on this interval):
$$ 1<\frac{\theta}{\sin\theta}<\frac{1}{\cos\theta}. $$Take reciprocals (and flip the inequalities):
$$ 1>\frac{\sin\theta}{\theta}>\cos\theta. $$Both outer functions head to 1 as \(\theta\to 0^+\) (\(\displaystyle\lim_{\theta\to 0^+}1=1\) and \(\displaystyle\lim_{\theta\to 0^+}\cos\theta=1\)). By the squeeze theorem,
$$ \lim_{\theta\to 0^+}\frac{\sin\theta}{\theta}=1. $$A symmetric argument (replacing \(\theta\) with \(-\theta\)) gives the left-hand limit, so
$$ \lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1. $$Example 2.3.13 uses this limit to establish a closely related and equally important fact.
Apply the squeeze theorem to evaluate \(\displaystyle\lim_{x\to 0}x\cos x.\)
Solution
Step 1 — Trap the function between two known bounds. Because \(-1\le \cos x\le 1\) for every \(x\), multiplying by \(x\) (and tracking the sign of \(x\)) gives
$$ -|x|\le x\cos x\le |x|. $$Step 2 — Check the outer limits agree.
$$ \lim_{x\to 0}\bigl(-|x|\bigr)=0=\lim_{x\to 0}|x|. $$Step 3 — Apply the squeeze theorem. Both outer functions head to 0 as \(x\to 0\), so the function trapped between them is squeezed to the same limit:
$$ \lim_{x\to 0}x\cos x=0. $$The graphs of \(f(x)=-|x|,\; g(x)=x\cos x,\) and \(h(x)=|x|\) are shown in Figure 2.28.
Figure 2.28 — The graphs of \(f(x),g(x),\) and \(h(x)\) are shown around the point \(x=0.\)
Answer: \(0.\)
Evaluate \(\displaystyle\lim_{\theta\to 0}\dfrac{1-\cos\theta}{\sin\theta}.\)
Solution
Step 1 — Rewrite as a product. Divide numerator and denominator by \(\theta\):
$$ \frac{1-\cos\theta}{\sin\theta}=\frac{(1-\cos\theta)/\theta}{(\sin\theta)/\theta}. $$Step 2 — Take limits of numerator and denominator. Numerator \(\to 0\) (Example 2.3.13), denominator \(\to 1\) (headline limit).
Step 3 — Apply the quotient law.
$$ \lim_{\theta\to 0}\frac{1-\cos\theta}{\sin\theta}=\frac{0}{1}=0. $$Answer: \(0.\)
Evaluate \(\displaystyle\lim_{\theta\to 0}\dfrac{1-\cos\theta}{\theta}.\)
Solution
Step 1 — Multiply by the conjugate \(1+\cos\theta\) so a Pythagorean identity surfaces in the numerator.
$$ \lim_{\theta\to 0}\frac{1-\cos\theta}{\theta}=\lim_{\theta\to 0}\frac{1-\cos\theta}{\theta}\cdot \frac{1+\cos\theta}{1+\cos\theta}=\lim_{\theta\to 0}\frac{1-\cos^2\theta}{\theta(1+\cos\theta)}. $$Step 2 — Replace \(1-\cos^2\theta\) with \(\sin^2\theta\).
$$ =\lim_{\theta\to 0}\frac{\sin^2\theta}{\theta(1+\cos\theta)}. $$Step 3 — Split into a familiar product.
$$ =\lim_{\theta\to 0}\frac{\sin\theta}{\theta}\cdot \frac{\sin\theta}{1+\cos\theta}. $$Step 4 — Take the limit of each factor. \(\dfrac{\sin\theta}{\theta}\to 1\) (the headline limit) and \(\dfrac{\sin\theta}{1+\cos\theta}\to \dfrac{0}{2}=0\):
$$ =1\cdot 0=0. $$Answer: \(\displaystyle\lim_{\theta\to 0}\dfrac{1-\cos\theta}{\theta}=0.\)
Student Project — Archimedes' Approach to the Area of a Circle
Some of the geometric formulas we take for granted today were first derived by methods that anticipate the methods of calculus. The Greek mathematician Archimedes (ca. 287–212 BCE) was particularly inventive, using polygons inscribed within a circle to approximate the circle's area as the number of sides grew. He never had the modern concept of a limit, but we can borrow it to see what his construction predicts.
The plan: approximate the area of a circle of radius \(r\) by inscribing a regular \(n\)-sided polygon, computing the polygon's area, and letting the vertex angle \(\theta\) of each isosceles triangle shrink toward zero (so \(n\to \infty\)). Carry out the following steps.
- Express the height \(h\) and the base \(b\) of the isosceles triangle in Figure 2.31 in terms of \(\theta\) and \(r\).
Figure 2.31
- Using the expressions from step 1, express the area of the isosceles triangle in terms of \(\theta\) and \(r\). (Hint: substitute \(\tfrac{1}{2}\sin\theta\) for \(\sin(\theta/2)\cos(\theta/2)\) in your expression.)
- If an \(n\)-sided regular polygon is inscribed in a circle of radius \(r\), find a relationship between \(\theta\) and \(n\), and solve for \(n\). Remember there are \(2\pi\) radians in a circle. (Use radians, not degrees.)
- Find an expression for the area of the \(n\)-sided polygon in terms of \(r\) and \(\theta\).
- To find a formula for the area of the circle, find the limit of the expression in step 4 as \(\theta\to 0.\) (Hint: \(\displaystyle\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta}=1.\))
The technique of estimating area by polygons reappears in Introduction to Integration.
Problem Set 2.3
In the following exercises, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).
Problem 1. \(\displaystyle\lim_{x\to 0}(4x^2-2x+3)\)
Solution
Step 1 — Apply the sum and difference laws. All three pieces are polynomials, so each piece has a limit:
$$ \lim_{x\to 0}(4x^2-2x+3)=\lim_{x\to 0}4x^2-\lim_{x\to 0}2x+\lim_{x\to 0}3. $$Step 2 — Apply the constant multiple and power laws.
$$ =4\bigl(\lim_{x\to 0}x\bigr)^2-2\lim_{x\to 0}x+\lim_{x\to 0}3. $$Step 3 — Apply the basic limit results and simplify.
$$ =4(0)^2-2(0)+3=3. $$Answer: \(\displaystyle\lim_{x\to 0}(4x^2-2x+3)=3.\)
Problem 2. \(\displaystyle\lim_{x\to 1}\dfrac{x^3+3x^2+5}{4-7x}\)
Solution
Step 1 — Check the denominator-limit. At \(x=1\), \(4-7(1)=-3\ne 0\), so the quotient law applies.
Step 2 — Apply the quotient law.
$$ \lim_{x\to 1}\frac{x^3+3x^2+5}{4-7x}=\frac{\displaystyle\lim_{x\to 1}(x^3+3x^2+5)}{\displaystyle\lim_{x\to 1}(4-7x)}. $$Step 3 — Evaluate each polynomial limit by direct substitution. Both pieces are polynomials, so we just plug in:
$$ =\frac{1^3+3(1)^2+5}{4-7(1)}=\frac{1+3+5}{-3}=\frac{9}{-3}=-3. $$Answer: \(-3.\)
Problem 3. \(\displaystyle\lim_{x\to -2}\sqrt{x^2-6x+3}\)
Solution
Step 1 — Apply the root law. The radicand-limit is the polynomial limit \(\lim_{x\to -2}(x^2-6x+3)\). We need it nonnegative to use the root law on a square root:
$$ \lim_{x\to -2}\sqrt{x^2-6x+3}=\sqrt{\lim_{x\to -2}(x^2-6x+3)}. $$Step 2 — Evaluate the polynomial limit.
$$ \lim_{x\to -2}(x^2-6x+3)=(-2)^2-6(-2)+3=4+12+3=19. $$Since \(19>0\), the root law applies.
Step 3 — Combine.
$$ \lim_{x\to -2}\sqrt{x^2-6x+3}=\sqrt{19}. $$Answer: \(\sqrt{19}.\)
Problem 4. \(\displaystyle\lim_{x\to -1}(9x+1)^2\)
In the following exercises, use direct substitution to evaluate each limit.
Solution
Step 1 — Apply the power law.
$$ \lim_{x\to -1}(9x+1)^2=\Bigl(\lim_{x\to -1}(9x+1)\Bigr)^2. $$Step 2 — Evaluate the inner polynomial limit.
$$ \lim_{x\to -1}(9x+1)=9(-1)+1=-8. $$Step 3 — Square the result.
$$ \lim_{x\to -1}(9x+1)^2=(-8)^2=64. $$Answer: \(64.\)
Problem 5. \(\displaystyle\lim_{x\to 7}x^2\)
Solution
Step 1 — Recognize a polynomial. \(x^2\) is a polynomial, so Theorem 2.3.3 lets us substitute directly.
Step 2 — Substitute.
$$ \lim_{x\to 7}x^2=7^2=49. $$Answer: \(49.\)
Problem 6. \(\displaystyle\lim_{x\to -2}(4x^2-1)\)
Solution
Step 1 — Polynomial limit; substitute.
$$ \lim_{x\to -2}(4x^2-1)=4(-2)^2-1=16-1=15. $$Answer: \(15.\)
Problem 7. \(\displaystyle\lim_{x\to 0}\dfrac{1}{1+\sin x}\)
Solution
Step 1 — Check continuity at \(x=0\). \(\sin x\) is continuous with \(\sin 0=0\), so \(1+\sin x\to 1\ne 0\). The quotient is continuous at \(x=0\); substitute.
Step 2 — Substitute.
$$ \lim_{x\to 0}\frac{1}{1+\sin x}=\frac{1}{1+\sin 0}=\frac{1}{1+0}=1. $$Answer: \(1.\)
Problem 8. \(\displaystyle\lim_{x\to 2}e^{2x-x^2}\)
Solution
Step 1 — The exponential is continuous. \(e^{u}\) is continuous in \(u\), and \(2x-x^2\) is a polynomial in \(x\), so the composition is continuous. Substitute directly.
Step 2 — Substitute.
$$ \lim_{x\to 2}e^{2x-x^2}=e^{2(2)-(2)^2}=e^{4-4}=e^0=1. $$Answer: \(1.\)
Problem 9. \(\displaystyle\lim_{x\to 1}\dfrac{2-7x}{x+6}\)
Solution
Step 1 — Check the denominator. At \(x=1\), \(x+6=7\ne 0\), so the rational function is continuous at \(x=1\).
Step 2 — Substitute.
$$ \lim_{x\to 1}\frac{2-7x}{x+6}=\frac{2-7(1)}{1+6}=\frac{-5}{7}=-\frac{5}{7}. $$Answer: \(-\dfrac{5}{7}.\)
Problem 10. \(\displaystyle\lim_{x\to 3}\ln e^{3x}\)
In the following exercises, use direct substitution to show that each limit leads to the indeterminate form \(0/0.\) Then, evaluate the limit.
Solution
Step 1 — Simplify using inverse functions. \(\ln\) and \(e^{(\cdot)}\) are inverses, so \(\ln e^{3x}=3x\) for every \(x\).
Step 2 — Substitute.
$$ \lim_{x\to 3}\ln e^{3x}=\lim_{x\to 3}3x=3(3)=9. $$Answer: \(9.\)
Problem 11. \(\displaystyle\lim_{x\to 4}\dfrac{x^2-16}{x-4}\)
Solution
Step 1 — Detect the indeterminate form. Substituting \(x=4\): numerator \(=16-16=0\), denominator \(=0\). Form is \(0/0\).
Step 2 — Factor the numerator. \(x^2-16=(x-4)(x+4)\), so
$$ \frac{x^2-16}{x-4}=\frac{(x-4)(x+4)}{x-4}=x+4 \quad\text{for } x\ne 4. $$Step 3 — Take the limit of the simplified expression.
$$ \lim_{x\to 4}(x+4)=4+4=8. $$Answer: \(8.\)
Problem 12. \(\displaystyle\lim_{x\to 2}\dfrac{x-2}{x^2-2x}\)
Solution
Step 1 — Detect the indeterminate form. At \(x=2\): numerator \(=0\); denominator \(=4-4=0\). Form is \(0/0\).
Step 2 — Factor the denominator and cancel.
$$ \frac{x-2}{x^2-2x}=\frac{x-2}{x(x-2)}=\frac{1}{x} \quad\text{for } x\ne 2. $$Step 3 — Take the limit of the simplified expression.
$$ \lim_{x\to 2}\frac{1}{x}=\frac{1}{2}. $$Answer: \(\dfrac{1}{2}.\)
Problem 13. \(\displaystyle\lim_{x\to 6}\dfrac{3x-18}{2x-12}\)
Solution
Step 1 — Detect the indeterminate form. At \(x=6\): numerator \(=18-18=0\); denominator \(=12-12=0\). Form is \(0/0\).
Step 2 — Factor common terms.
$$ \frac{3x-18}{2x-12}=\frac{3(x-6)}{2(x-6)}=\frac{3}{2} \quad\text{for } x\ne 6. $$Step 3 — Take the limit. A constant function has limit equal to that constant:
$$ \lim_{x\to 6}\frac{3}{2}=\frac{3}{2}. $$Answer: \(\dfrac{3}{2}.\)
Problem 14. \(\displaystyle\lim_{h\to 0}\dfrac{(1+h)^2-1}{h}\)
Solution
Step 1 — Detect the indeterminate form. At \(h=0\): numerator \(=1-1=0\); denominator \(=0\). Form is \(0/0\).
Step 2 — Expand the numerator.
$$ (1+h)^2-1=1+2h+h^2-1=2h+h^2. $$Step 3 — Cancel \(h\).
$$ \frac{2h+h^2}{h}=\frac{h(2+h)}{h}=2+h \quad\text{for } h\ne 0. $$Step 4 — Take the limit.
$$ \lim_{h\to 0}(2+h)=2. $$Answer: \(2.\)
Problem 15. \(\displaystyle\lim_{t\to 9}\dfrac{t-9}{\sqrt{t}-3}\)
Solution
Step 1 — Detect the indeterminate form. At \(t=9\): numerator \(=0\); denominator \(=\sqrt{9}-3=0\). Form is \(0/0\).
Step 2 — Factor the numerator as a difference of squares. \(t-9=(\sqrt{t})^2-3^2=(\sqrt{t}-3)(\sqrt{t}+3)\):
$$ \frac{t-9}{\sqrt{t}-3}=\frac{(\sqrt{t}-3)(\sqrt{t}+3)}{\sqrt{t}-3}=\sqrt{t}+3 \quad\text{for } t\ne 9. $$Step 3 — Take the limit. \(\sqrt{t}+3\) is continuous at \(t=9\):
$$ \lim_{t\to 9}(\sqrt{t}+3)=\sqrt{9}+3=6. $$Answer: \(6.\)
Problem 16. \(\displaystyle\lim_{h\to 0}\dfrac{\frac{1}{a+h}-\frac{1}{a}}{h},\) where \(a\) is a nonzero real-valued constant
Solution
Step 1 — Detect the indeterminate form. At \(h=0\): numerator \(=\tfrac{1}{a}-\tfrac{1}{a}=0\); denominator \(=0\). Form is \(0/0\). The expression is a complex fraction.
Step 2 — Combine the inner fractions in the numerator. Common denominator \(a(a+h)\):
$$ \frac{1}{a+h}-\frac{1}{a}=\frac{a-(a+h)}{a(a+h)}=\frac{-h}{a(a+h)}. $$Step 3 — Divide by \(h\) (i.e., multiply by \(1/h\)).
$$ \frac{\frac{-h}{a(a+h)}}{h}=\frac{-h}{a(a+h)\cdot h}=\frac{-1}{a(a+h)} \quad\text{for } h\ne 0. $$Step 4 — Take the limit. \(\dfrac{-1}{a(a+h)}\) is continuous at \(h=0\) (\(a\ne 0\)):
$$ \lim_{h\to 0}\frac{-1}{a(a+h)}=\frac{-1}{a\cdot a}=-\frac{1}{a^2}. $$Answer: \(-\dfrac{1}{a^2}.\)
Problem 17. \(\displaystyle\lim_{\theta\to \pi}\dfrac{\sin\theta}{\tan\theta}\)
Solution
Step 1 — Detect the indeterminate form. At \(\theta=\pi\): \(\sin\pi=0\) and \(\tan\pi=0\). Form is \(0/0\).
Step 2 — Simplify using \(\tan\theta=\sin\theta/\cos\theta\).
$$ \frac{\sin\theta}{\tan\theta}=\frac{\sin\theta}{\sin\theta/\cos\theta}=\cos\theta \quad\text{whenever } \sin\theta\ne 0. $$The simplification is valid on a deleted neighborhood of \(\pi\).
Step 3 — Take the limit. \(\cos\theta\) is continuous everywhere:
$$ \lim_{\theta\to \pi}\cos\theta=\cos\pi=-1. $$Answer: \(-1.\)
Problem 18. \(\displaystyle\lim_{x\to 1}\dfrac{x^3-1}{x^2-1}\)
Solution
Step 1 — Detect the indeterminate form. At \(x=1\): numerator \(=0\); denominator \(=0\). Form is \(0/0\).
Step 2 — Factor and cancel. \(x^3-1=(x-1)(x^2+x+1)\) and \(x^2-1=(x-1)(x+1)\), so
$$ \frac{x^3-1}{x^2-1}=\frac{(x-1)(x^2+x+1)}{(x-1)(x+1)}=\frac{x^2+x+1}{x+1} \quad\text{for } x\ne 1. $$Step 3 — Substitute.
$$ \lim_{x\to 1}\frac{x^2+x+1}{x+1}=\frac{1+1+1}{1+1}=\frac{3}{2}. $$Answer: \(\dfrac{3}{2}.\)
Problem 19. \(\displaystyle\lim_{x\to 1/2}\dfrac{2x^2+3x-2}{2x-1}\)
Solution
Step 1 — Detect the indeterminate form. At \(x=\tfrac{1}{2}\): numerator \(=2(\tfrac{1}{4})+\tfrac{3}{2}-2=\tfrac{1}{2}+\tfrac{3}{2}-2=0\); denominator \(=0\). Form is \(0/0\).
Step 2 — Factor the numerator. Try \(2x^2+3x-2=(2x-1)(x+2)\). Check: \((2x-1)(x+2)=2x^2+4x-x-2=2x^2+3x-2\). Good.
$$ \frac{2x^2+3x-2}{2x-1}=\frac{(2x-1)(x+2)}{2x-1}=x+2 \quad\text{for } x\ne \tfrac{1}{2}. $$Step 3 — Substitute.
$$ \lim_{x\to 1/2}(x+2)=\tfrac{1}{2}+2=\tfrac{5}{2}. $$Answer: \(\dfrac{5}{2}.\)
Problem 20. \(\displaystyle\lim_{x\to -3}\dfrac{\sqrt{x+4}-1}{x+3}\)
In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of Example 2.3.11 to simplify the function to help determine the limit.
Solution
Step 1 — Detect the indeterminate form. At \(x=-3\): numerator \(=\sqrt{1}-1=0\); denominator \(=0\). Form is \(0/0\).
Step 2 — Multiply by the conjugate \(\sqrt{x+4}+1\).
$$ \frac{\sqrt{x+4}-1}{x+3}\cdot \frac{\sqrt{x+4}+1}{\sqrt{x+4}+1}=\frac{(x+4)-1}{(x+3)(\sqrt{x+4}+1)}=\frac{x+3}{(x+3)(\sqrt{x+4}+1)}. $$Step 3 — Cancel \((x+3)\).
$$ =\frac{1}{\sqrt{x+4}+1} \quad\text{for } x\ne -3. $$Step 4 — Take the limit. The radicand \(x+4\to 1>0\), so the root law applies:
$$ \lim_{x\to -3}\frac{1}{\sqrt{x+4}+1}=\frac{1}{\sqrt{1}+1}=\frac{1}{2}. $$Answer: \(\dfrac{1}{2}.\)
Problem 21. \(\displaystyle\lim_{x\to -2^-}\dfrac{2x^2+7x-4}{x^2+x-2}\)
Solution
Step 1 — Detect the form. At \(x=-2\): numerator \(=2(4)+(-14)-4=8-14-4=-10\); denominator \(=4-2-2=0\). Form is \(-10/0\) — magnitude explodes; we need the sign.
Step 2 — Factor and split into a product. Factor the numerator \(2x^2+7x-4=(2x-1)(x+4)\) and the denominator \(x^2+x-2=(x-1)(x+2)\):
$$ \frac{2x^2+7x-4}{x^2+x-2}=\frac{(2x-1)(x+4)}{(x-1)(x+2)}=\frac{(2x-1)(x+4)}{x-1}\cdot \frac{1}{x+2}. $$Step 3 — Take the finite factor's limit. At \(x=-2\): \(\dfrac{(2(-2)-1)((-2)+4)}{-2-1}=\dfrac{(-5)(2)}{-3}=\dfrac{-10}{-3}=\dfrac{10}{3}>0.\)
Step 4 — Determine the sign of \(1/(x+2)\) as \(x\to -2^-\). For \(x<-2\), \(x+2<0\), so \(x+2\to 0^-\) and \(\dfrac{1}{x+2}\to -\infty.\)
Step 5 — Combine. Positive finite \(\times\) negative infinity = negative infinity.
Answer: \(-\infty.\)
Problem 22. \(\displaystyle\lim_{x\to -2^+}\dfrac{2x^2+7x-4}{x^2+x-2}\)
Solution
Step 1 — Reuse the factoring from 2.3.21.
$$ \frac{2x^2+7x-4}{x^2+x-2}=\frac{(2x-1)(x+4)}{x-1}\cdot \frac{1}{x+2}. $$The finite factor still tends to \(\dfrac{10}{3}>0\) as \(x\to -2\).
Step 2 — Determine the sign of \(1/(x+2)\) as \(x\to -2^+\). For \(x>-2\), \(x+2>0\), so \(x+2\to 0^+\) and \(\dfrac{1}{x+2}\to +\infty.\)
Step 3 — Combine. Positive finite \(\times\) positive infinity = positive infinity.
Answer: \(+\infty.\)
Problem 23. \(\displaystyle\lim_{x\to 1^-}\dfrac{2x^2+7x-4}{x^2+x-2}\)
Solution
Step 1 — Detect the form. At \(x=1\): numerator \(=2+7-4=5\); denominator \(=1+1-2=0\). Form is \(5/0\) — magnitude explodes.
Step 2 — Factor and split. Using the same factoring as 2.3.21,
$$ \frac{2x^2+7x-4}{x^2+x-2}=\frac{(2x-1)(x+4)}{x+2}\cdot \frac{1}{x-1}. $$Step 3 — Take the finite factor's limit at \(x=1\). \(\dfrac{(2(1)-1)(1+4)}{1+2}=\dfrac{(1)(5)}{3}=\dfrac{5}{3}>0.\)
Step 4 — Determine the sign of \(1/(x-1)\) as \(x\to 1^-\). For \(x<1\), \(x-1<0\), so \(x-1\to 0^-\) and \(\dfrac{1}{x-1}\to -\infty.\)
Step 5 — Combine. Positive finite \(\times\) negative infinity = negative infinity.
Answer: \(-\infty.\)
Problem 24. \(\displaystyle\lim_{x\to 1^+}\dfrac{2x^2+7x-4}{x^2+x-2}\)
In the following exercises, assume that \(\displaystyle\lim_{x\to 6}f(x)=4,\ \lim_{x\to 6}g(x)=9,\) and \(\displaystyle\lim_{x\to 6}h(x)=6.\) Use these three facts and the limit laws to evaluate each limit.
Solution
Step 1 — Reuse the factoring from 2.3.23.
$$ \frac{2x^2+7x-4}{x^2+x-2}=\frac{(2x-1)(x+4)}{x+2}\cdot \frac{1}{x-1}. $$The finite factor tends to \(\dfrac{5}{3}>0\) as \(x\to 1\).
Step 2 — Sign of \(1/(x-1)\) as \(x\to 1^+\). For \(x>1\), \(x-1>0\), so \(x-1\to 0^+\) and \(\dfrac{1}{x-1}\to +\infty.\)
Step 3 — Combine. Positive finite \(\times\) positive infinity = positive infinity.
Answer: \(+\infty.\)
Problem 25. \(\displaystyle\lim_{x\to 6}2f(x)g(x)\)
Solution
Step 1 — Apply the constant multiple and product laws.
$$ \lim_{x\to 6}2f(x)g(x)=2\cdot \lim_{x\to 6}f(x)\cdot \lim_{x\to 6}g(x). $$Step 2 — Substitute the given limits.
$$ =2\cdot 4\cdot 9=72. $$Answer: \(72.\)
Problem 26. \(\displaystyle\lim_{x\to 6}\dfrac{g(x)-1}{f(x)}\)
Solution
Step 1 — Apply the quotient law. The denominator-limit is \(4\ne 0\), so the quotient law applies:
$$ \lim_{x\to 6}\frac{g(x)-1}{f(x)}=\frac{\lim_{x\to 6}g(x)-\lim_{x\to 6}1}{\lim_{x\to 6}f(x)}. $$Step 2 — Substitute.
$$ =\frac{9-1}{4}=\frac{8}{4}=2. $$Answer: \(2.\)
Problem 27. \(\displaystyle\lim_{x\to 6}\!\left(f(x)+\tfrac{1}{3}g(x)\right)\)
Solution
Step 1 — Apply the sum and constant multiple laws.
$$ \lim_{x\to 6}\!\left(f(x)+\tfrac{1}{3}g(x)\right)=\lim_{x\to 6}f(x)+\tfrac{1}{3}\lim_{x\to 6}g(x). $$Step 2 — Substitute.
$$ =4+\tfrac{1}{3}(9)=4+3=7. $$Answer: \(7.\)
Problem 28. \(\displaystyle\lim_{x\to 6}\dfrac{(h(x))^3}{2}\)
Solution
Step 1 — Apply the constant multiple and power laws.
$$ \lim_{x\to 6}\frac{(h(x))^3}{2}=\frac{1}{2}\Bigl(\lim_{x\to 6}h(x)\Bigr)^3. $$Step 2 — Substitute.
$$ =\frac{1}{2}(6)^3=\frac{216}{2}=108. $$Answer: \(108.\)
Problem 29. \(\displaystyle\lim_{x\to 6}\sqrt{g(x)-f(x)}\)
Solution
Step 1 — Apply the root and difference laws. The radicand-limit is \(9-4=5\ge 0\), so the (even-index) root law applies:
$$ \lim_{x\to 6}\sqrt{g(x)-f(x)}=\sqrt{\lim_{x\to 6}g(x)-\lim_{x\to 6}f(x)}. $$Step 2 — Substitute.
$$ =\sqrt{9-4}=\sqrt{5}. $$Answer: \(\sqrt{5}.\)
Problem 30. \(\displaystyle\lim_{x\to 6}x\cdot h(x)\)
Solution
Step 1 — Apply the product law.
$$ \lim_{x\to 6}x\cdot h(x)=\lim_{x\to 6}x\cdot \lim_{x\to 6}h(x). $$Step 2 — Substitute. The basic limit result gives \(\lim_{x\to 6}x=6\):
$$ =6\cdot 6=36. $$Answer: \(36.\)
Problem 31. \(\displaystyle\lim_{x\to 6}\bigl[(x+1)\cdot f(x)\bigr]\)
Solution
Step 1 — Apply the product and sum laws.
$$ \lim_{x\to 6}\bigl[(x+1)f(x)\bigr]=\Bigl(\lim_{x\to 6}(x+1)\Bigr)\Bigl(\lim_{x\to 6}f(x)\Bigr). $$Step 2 — Substitute.
$$ =(6+1)(4)=7\cdot 4=28. $$Answer: \(28.\)
Problem 32. \(\displaystyle\lim_{x\to 6}\bigl(f(x)\cdot g(x)-h(x)\bigr)\)
[T] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.
Solution
Step 1 — Apply the difference and product laws.
$$ \lim_{x\to 6}\bigl(f(x)g(x)-h(x)\bigr)=\Bigl(\lim_{x\to 6}f(x)\Bigr)\Bigl(\lim_{x\to 6}g(x)\Bigr)-\lim_{x\to 6}h(x). $$Step 2 — Substitute.
$$ =(4)(9)-6=36-6=30. $$Answer: \(30.\)
Problem 33. \(f(x)=\begin{cases}x^2,& x\le 3 \\ x+4,& x>3\end{cases}\)
a) \(\displaystyle\lim_{x\to 3^-}f(x)\)
b) \(\displaystyle\lim_{x\to 3^+}f(x)\)
Solution
Step 1 — Pick the relevant branch on each side. For \(x<3\), \(f(x)=x^2\). For \(x>3\), \(f(x)=x+4\). The single value \(f(3)=9\) (from the top branch, since \(x\le 3\)) does not affect either one-sided limit.
Step 2 — Left-hand limit. Use \(x^2\):
$$ \lim_{x\to 3^-}f(x)=\lim_{x\to 3^-}x^2=3^2=9. $$Step 3 — Right-hand limit. Use \(x+4\):
$$ \lim_{x\to 3^+}f(x)=\lim_{x\to 3^+}(x+4)=3+4=7. $$Note on the graphing calculator. The plot would show a parabola \(y=x^2\) climbing to a closed point at \((3,9)\), then a line \(y=x+4\) jumping down to an open point at \((3,7)\) and continuing upward.
Answer: a) \(9\); b) \(7\).
Problem 34. \(g(x)=\begin{cases}x^3-1,& x\le 0 \\ 1,& x>0\end{cases}\)
a) \(\displaystyle\lim_{x\to 0^-}g(x)\)
b) \(\displaystyle\lim_{x\to 0^+}g(x)\)
Solution
Step 1 — Left-hand limit. For \(x<0\), \(g(x)=x^3-1\):
$$ \lim_{x\to 0^-}g(x)=\lim_{x\to 0^-}(x^3-1)=0-1=-1. $$Step 2 — Right-hand limit. For \(x>0\), \(g(x)=1\):
$$ \lim_{x\to 0^+}g(x)=\lim_{x\to 0^+}1=1. $$Note on the graphing calculator. The plot would show a cubic curve rising through \((-1,-2)\) up to a closed point at \((0,-1)\), then a horizontal segment at height 1 starting at an open point at \((0,1)\) and extending to the right.
Answer: a) \(-1\); b) \(1\).
Problem 35. \(h(x)=\begin{cases}x^2-2x+1,& x<2 \\ 3-x,& x\ge 2\end{cases}\)
a) \(\displaystyle\lim_{x\to 2^-}h(x)\)
b) \(\displaystyle\lim_{x\to 2^+}h(x)\)
In the following exercises, use the graphs below and the limit laws to evaluate each limit.
Solution
Step 1 — Left-hand limit. For \(x<2\), \(h(x)=x^2-2x+1=(x-1)^2\):
$$ \lim_{x\to 2^-}h(x)=\lim_{x\to 2^-}(x^2-2x+1)=4-4+1=1. $$Step 2 — Right-hand limit. For \(x\ge 2\), \(h(x)=3-x\):
$$ \lim_{x\to 2^+}h(x)=\lim_{x\to 2^+}(3-x)=3-2=1. $$Step 3 — Comment. Both one-sided limits agree at \(1\), so the two-sided limit \(\lim_{x\to 2}h(x)\) also equals 1. The two branches meet at the point \((2,1)\) on the graph.
Answer: a) \(1\); b) \(1\).
Problem 36. \(\displaystyle\lim_{x\to -3^+}(f(x)+g(x))\)
Solution
Step 1 — Apply the sum law. Provided both pieces have one-sided limits at \(x=-3^+\),
$$ \lim_{x\to -3^+}(f(x)+g(x))=\lim_{x\to -3^+}f(x)+\lim_{x\to -3^+}g(x). $$Step 2 — Read the one-sided values off the graph. From the figure, \(\lim_{x\to -3^+}f(x)=1\) and \(\lim_{x\to -3^+}g(x)=1\).
Step 3 — Combine.
$$ \lim_{x\to -3^+}(f(x)+g(x))=1+1=2. $$Answer: \(2.\)
Problem 37. \(\displaystyle\lim_{x\to -3^-}(f(x)-3g(x))\)
Solution
Step 1 — Apply the difference and constant multiple laws.
$$ \lim_{x\to -3^-}(f(x)-3g(x))=\lim_{x\to -3^-}f(x)-3\lim_{x\to -3^-}g(x). $$Step 2 — Read the one-sided values off the graph. From the figure, \(\lim_{x\to -3^-}f(x)=4\) and \(\lim_{x\to -3^-}g(x)=3\).
Step 3 — Combine.
$$ \lim_{x\to -3^-}(f(x)-3g(x))=4-3(3)=4-9=-5. $$Answer: \(-5.\)
Problem 38. \(\displaystyle\lim_{x\to 0}\dfrac{f(x)g(x)}{3}\)
Solution
Step 1 — Apply the constant multiple and product laws.
$$ \lim_{x\to 0}\frac{f(x)g(x)}{3}=\frac{1}{3}\Bigl(\lim_{x\to 0}f(x)\Bigr)\Bigl(\lim_{x\to 0}g(x)\Bigr). $$Step 2 — Read the values off the graph. From the figure, \(\lim_{x\to 0}f(x)=3\) and \(\lim_{x\to 0}g(x)=-3\).
Step 3 — Combine.
$$ =\frac{1}{3}(3)(-3)=-3. $$Answer: \(-3.\)
Problem 39. \(\displaystyle\lim_{x\to -5}\dfrac{2+g(x)}{f(x)}\)
Solution
Step 1 — Apply the quotient and sum laws. First check the denominator-limit is nonzero. From the figure, \(\lim_{x\to -5}f(x)=2\ne 0\), so
$$ \lim_{x\to -5}\frac{2+g(x)}{f(x)}=\frac{2+\lim_{x\to -5}g(x)}{\lim_{x\to -5}f(x)}. $$Step 2 — Read the limit of \(g\). From the figure, \(\lim_{x\to -5}g(x)=2\).
Step 3 — Combine.
$$ =\frac{2+2}{2}=\frac{4}{2}=2. $$Answer: \(2.\)
Problem 40. \(\displaystyle\lim_{x\to 1}(f(x))^2\)
Solution
Step 1 — Apply the power law.
$$ \lim_{x\to 1}(f(x))^2=\Bigl(\lim_{x\to 1}f(x)\Bigr)^2. $$Step 2 — Read \(f\) at \(x=1\) from the graph. From the figure, \(\lim_{x\to 1}f(x)=-2\).
Step 3 — Square.
$$ \lim_{x\to 1}(f(x))^2=(-2)^2=4. $$Answer: \(4.\)
Problem 41. \(\displaystyle\lim_{x\to 1}\sqrt[3]{f(x)-g(x)}\)
Solution
Step 1 — Apply the cube-root and difference laws. The cube root law applies for any real value (odd index):
$$ \lim_{x\to 1}\sqrt[3]{f(x)-g(x)}=\sqrt[3]{\lim_{x\to 1}f(x)-\lim_{x\to 1}g(x)}. $$Step 2 — Read the values off the graph. From the figure, \(\lim_{x\to 1}f(x)=-2\) and \(\lim_{x\to 1}g(x)=5\).
Step 3 — Combine.
$$ =\sqrt[3]{-2-5}=\sqrt[3]{-7}=-\sqrt[3]{7}. $$Answer: \(-\sqrt[3]{7}.\)
Problem 42. \(\displaystyle\lim_{x\to -7}(x\cdot g(x))\)
Solution
Step 1 — Apply the product law.
$$ \lim_{x\to -7}(x\cdot g(x))=\Bigl(\lim_{x\to -7}x\Bigr)\Bigl(\lim_{x\to -7}g(x)\Bigr). $$Step 2 — Evaluate each factor. The basic limit result gives \(\lim_{x\to -7}x=-7\). From the figure, \(\lim_{x\to -7}g(x)=1\).
Step 3 — Combine.
$$ \lim_{x\to -7}(x\cdot g(x))=(-7)(1)=-7. $$Answer: \(-7.\)
Problem 43. \(\displaystyle\lim_{x\to -9}[x\cdot f(x)+2\cdot g(x)]\)
For the following problems, evaluate the limit using the squeeze theorem. Use a calculator to graph the functions \(f(x),\;g(x),\) and \(h(x)\) when possible.
Solution
Step 1 — Apply the sum, product, and constant multiple laws.
$$ \lim_{x\to -9}[xf(x)+2g(x)]=\Bigl(\lim_{x\to -9}x\Bigr)\Bigl(\lim_{x\to -9}f(x)\Bigr)+2\lim_{x\to -9}g(x). $$Step 2 — Evaluate each piece. The basic limit result gives \(\lim_{x\to -9}x=-9\). From the figure, \(\lim_{x\to -9}f(x)=6\) and \(\lim_{x\to -9}g(x)=4\).
Step 3 — Combine.
$$ =(-9)(6)+2(4)=-54+8=-46. $$Answer: \(-46.\)
Problem 44. [T] True or False? If \(2x-1\le g(x)\le x^2-2x+3,\) then \(\displaystyle\lim_{x\to 2}g(x)=0.\)
Solution
Step 1 — Check the outer functions at \(x=2\). Lower bound: \(2(2)-1=3\). Upper bound: \(2^2-2(2)+3=4-4+3=3.\)
Step 2 — Apply the squeeze theorem. Both outer functions have limit \(3\) at \(x=2\), so any \(g\) trapped between them is squeezed to \(3\):
$$ \lim_{x\to 2}g(x)=3, $$not \(0\). The claim that the limit equals 0 is false.
Note on the graphing calculator. Plotting \(y=2x-1\) and \(y=x^2-2x+3\) shows the two curves crossing at \((2,3)\); a graph of \(g\) (if known) would be visibly trapped between them and pinched to that crossing point.
Answer: False. The squeeze theorem forces \(\lim_{x\to 2}g(x)=3.\)
Problem 45. [T] \(\displaystyle\lim_{\theta\to 0}\theta^2\cos\!\left(\dfrac{1}{\theta}\right)\)
Solution
Step 1 — Trap the function between two known bounds. Because \(-1\le \cos(1/\theta)\le 1\) for every \(\theta\ne 0\), multiplying by \(\theta^2\ge 0\) preserves the inequality:
$$ -\theta^2\le \theta^2\cos\!\frac{1}{\theta}\le \theta^2. $$Step 2 — Check the outer limits agree.
$$ \lim_{\theta\to 0}(-\theta^2)=0=\lim_{\theta\to 0}\theta^2. $$Step 3 — Apply the squeeze theorem.
$$ \lim_{\theta\to 0}\theta^2\cos\!\frac{1}{\theta}=0. $$Note on the graphing calculator. A plot of the three functions near 0 would show \(y=\theta^2\cos(1/\theta)\) wildly oscillating but firmly pinned inside the parabolic envelopes \(y=\pm\theta^2\), all three collapsing to the origin.
Answer: \(0.\)
Problem 46. \(\displaystyle\lim_{x\to 0}f(x),\) where \(f(x)=\begin{cases}0,& x \text{ rational} \\ x^2,& x \text{ irrational}\end{cases}\)
Solution
Step 1 — Trap \(f\) between two simple functions. Whether \(x\) is rational or irrational, \(0\le x^2\); and \(f(x)\) is either \(0\) or \(x^2\), so
$$ 0\le f(x)\le x^2 \quad\text{for every } x. $$Step 2 — Check the outer limits agree.
$$ \lim_{x\to 0}0=0=\lim_{x\to 0}x^2. $$Step 3 — Apply the squeeze theorem.
$$ \lim_{x\to 0}f(x)=0. $$Answer: \(0.\) (The limit exists even though \(f\) is wildly discontinuous away from \(0\) — the trap closes only at the single point \(x=0\).)
Problem 47. [T] In physics, the magnitude of an electric field generated by a point charge at a distance \(r\) in vacuum is governed by Coulomb's law: \(E(r)=\dfrac{q}{4\pi\varepsilon_0 r^2},\) where \(E\) is the magnitude of the electric field, \(q\) is the charge of the particle, \(r\) is the distance between the particle and where the field is measured, and \(\dfrac{1}{4\pi\varepsilon_0}\) is Coulomb's constant: \(8.988\times 10^9\;\text{N}\cdot\text{m}^2/\text{C}^2.\)
a) Use a graphing calculator to graph \(E(r)\) given that the charge of the particle is \(q=10^{-10}.\)
b) Evaluate \(\displaystyle\lim_{r\to 0^+}E(r).\) What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?
Solution
Step 1 — Write \(E(r)\) explicitly for the given charge. Multiply Coulomb's constant by \(q=10^{-10}\):
$$ E(r)=\frac{q}{4\pi\varepsilon_0 r^2}=\frac{(8.988\times 10^9)(10^{-10})}{r^2}=\frac{0.8988}{r^2}\;\text{N/C}. $$Step 2 — Read the form of the limit. As \(r\to 0^+\), the numerator stays at the positive constant \(0.8988\), and the denominator \(r^2\to 0^+\). Form is (positive)/(positive zero) — the magnitude blows up positively:
$$ \lim_{r\to 0^+}E(r)=\lim_{r\to 0^+}\frac{0.8988}{r^2}=+\infty. $$Step 3 — Physical meaning. The model predicts that the electric field magnitude grows without bound as you approach the point charge. This is the mathematical singularity in the idealized point-charge model — it is not physically realized, because real charges have finite extent and quantum effects take over at small distances.
Step 4 — Why right-hand only? Distance \(r\) is a nonnegative quantity, so values \(r<0\) have no physical meaning. The only side that exists on the domain is \(r\to 0^+\).
Note on the graphing calculator. A plot of \(E(r)=0.8988/r^2\) would show a curve hugging the horizontal axis far from \(r=0\) and rising steeply upward, asymptotically vertical, as \(r\to 0^+\).
Answer: \(\displaystyle\lim_{r\to 0^+}E(r)=+\infty.\) Mathematically the field diverges at the charge; this is an artifact of the point-charge idealization, not a physically observed value. Only the right-hand limit is meaningful because \(r\) is a distance and must be positive.
Problem 48. [T] The density of an object is given by its mass divided by its volume: \(\rho=m/V.\)
a) Use a calculator to plot the volume as a function of density \((V=m/\rho),\) assuming you are examining something of mass 8 kg \((m=8).\)
b) Evaluate \(\displaystyle\lim_{\rho\to 0^+}V(\rho)\) and explain the physical meaning.
Solution
Step 1 — Write the volume function for \(m=8\). Solving \(\rho=m/V\) for \(V\) gives \(V=m/\rho\), so
$$ V(\rho)=\frac{8}{\rho}. $$Step 2 — Read the form of the limit. As \(\rho\to 0^+\), the numerator stays at \(8>0\), and the denominator \(\to 0^+\). Form is (positive)/(positive zero) — the magnitude blows up positively:
$$ \lim_{\rho\to 0^+}V(\rho)=\lim_{\rho\to 0^+}\frac{8}{\rho}=+\infty. $$Step 3 — Physical meaning. Holding mass fixed at \(8\) kg and squeezing density toward zero forces volume to grow without bound. In words: a very-low-density object with a fixed mass must occupy a very large volume. Pushing density all the way to zero would require infinite volume — physically unreachable, but the trend is real (think of gases at very low pressure expanding to fill larger and larger containers).
Step 4 — Why right-hand only? Density is a positive physical quantity, so only \(\rho\to 0^+\) makes physical sense.
Note on the graphing calculator. A plot of \(V=8/\rho\) is a rectangular hyperbola in the first quadrant: it falls toward the \(\rho\)-axis as density grows and climbs steeply upward as density approaches 0.
Answer: \(\displaystyle\lim_{\rho\to 0^+}V(\rho)=+\infty.\) Holding mass fixed, the volume needed to realize a vanishing density grows without bound.
Key Terms
limit laws — the individual algebraic properties that allow the limit symbol to be distributed across sums, differences, products, quotients, powers, and roots.
Sum law for limits — \(\displaystyle\lim_{x\to a}(f(x)+g(x))=\lim_{x\to a}f(x)+\lim_{x\to a}g(x).\)
Difference law for limits — \(\displaystyle\lim_{x\to a}(f(x)-g(x))=\lim_{x\to a}f(x)-\lim_{x\to a}g(x).\)
Constant multiple law for limits — \(\displaystyle\lim_{x\to a}cf(x)=c\cdot \lim_{x\to a}f(x).\)
Product law for limits — \(\displaystyle\lim_{x\to a}(f(x)g(x))=\lim_{x\to a}f(x)\cdot \lim_{x\to a}g(x).\)
Quotient law for limits — \(\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}\) for nonzero denominator-limit.
Power law for limits — \(\displaystyle\lim_{x\to a}(f(x))^n=\bigl(\lim_{x\to a}f(x)\bigr)^n\) for positive integer \(n\).
Root law for limits — \(\displaystyle\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\to a}f(x)}\), with sign restrictions when \(n\) is even.
squeeze theorem — if \(f(x)\le g(x)\le h(x)\) on a neighborhood of \(a\) and \(\lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L\), then \(\lim_{x\to a}g(x)=L\).
Archimedes — Greek mathematician (ca. 287–212 BCE) who approximated the area of a circle with inscribed regular polygons.