2.5 The Precise Definition of a Limit
Learning Objectives
- Describe the epsilon–delta definition of a limit.
- Apply the epsilon–delta definition to find the limit of a function.
- Describe the epsilon–delta definitions of one-sided limits and infinite limits.
- Use the epsilon–delta definition to prove the limit laws.
By now you've climbed a real ladder. You started with an informal "limits feel like…" picture in the chapter introduction, sharpened it into an intuitive idea, and used that intuition to compute limits, classify discontinuities, and apply the Intermediate Value Theorem. In this section we do one more thing: we replace "feels like" with precise mathematical language. The result is the ε–δ definition of a limit, which is famously the most technical definition you meet in early calculus — and is also the one that unlocks every proof later in the course. The payoff is worth the effort.
2.5.1 Quantifying Closeness
Before we can write down the formal definition, we need a way to talk about closeness — how close two numbers are to each other — using just arithmetic, not pictures.
Recall that the distance between two points \(a\) and \(b\) on a number line is \(|a-b|\). So when we say "\(f(x)\) is within \(\varepsilon\) of \(L\)" we really mean \(|f(x)-L|<\varepsilon\). Two language ↔ symbol pairs we'll lean on:
- The statement \(|f(x)-L|<\varepsilon\) reads as: the distance between \(f(x)\) and \(L\) is less than \(\varepsilon\).
- The statement \(0<|x-a|<\delta\) reads as: \(x\ne a\) and the distance between \(x\) and \(a\) is less than \(\delta\). The lower bound \(0<\) is doing real work — it explicitly excludes the case \(x=a\), because the limit is about the approach, not the value at the point.
Inside \(|x-a|<\delta\) sits the more useful unwrapped form \(a-\delta < x < a+\delta\) — a width-\(2\delta\) interval centered on \(a\). When you're trying to picture an ε–δ proof, you're usually looking at TWO horizontal strips and TWO vertical strips: an ε-band around \(y=L\) and a δ-band around \(x=a\). The absolute-value form is compact for writing; the inequality form is the picture.
Two equivalences make this swap explicit:
- \(|f(x)-L|<\varepsilon\) is equivalent to \(L-\varepsilon<f(x)<L+\varepsilon\).
- \(0<|x-a|<\delta\) is equivalent to \(a-\delta<x<a+\delta\) with \(x\ne a\).
With this, we're ready for the formal definition.
Let \(f(x)\) be defined for all \(x\ne a\) over an open interval containing \(a\). Let \(L\) be a real number. Then
$$ \lim_{x\to a}f(x)=L $$if, for every \(\varepsilon>0\), there exists a \(\delta>0\) such that if \(0<|x-a|<\delta\), then \(|f(x)-L|<\varepsilon\).
This sentence has three structural pieces — a universal quantifier ("for every \(\varepsilon>0\)"), an existential quantifier ("there exists a \(\delta>0\)"), and a conditional statement ("if \(0<|x-a|<\delta\), then \(|f(x)-L|<\varepsilon\)"). The order matters: the quantifiers run outside-in, so for any \(\varepsilon\) the opponent picks, we have to be able to find a corresponding \(\delta\). Small ε, smaller δ.
Figure 2.39 shows the picture: three panels with successively smaller \(\varepsilon\). In each panel, we find a corresponding \(\delta\) so that the rectangle around \((a, L)\) made by the δ-band on the \(x\)-axis and the ε-band on the \(y\)-axis traps the graph. As ε shrinks, δ shrinks with it.
Figure 2.39 — These graphs show possible values of \(\delta\), given successively smaller choices of \(\varepsilon\).
The next example shows exactly how to use this definition to prove a limit. Pay attention to the structure — it's a recipe that repeats.
Complete the proof that \(\lim_{x\to 2}(3x-2)=4\) by filling in the blanks.
Let \(\varepsilon>0\). Choose \(\delta=\;\rule{1cm}{0.4pt}\;\). Assume \(0<|x-\;\rule{0.6cm}{0.4pt}\;|<\;\rule{0.6cm}{0.4pt}\;\). Thus, \(|\;\rule{1cm}{0.4pt}\;-\;\rule{0.6cm}{0.4pt}\;|=\;\rule{2cm}{0.4pt}\;<\varepsilon\). Therefore, \(\lim_{x\to 2}(3x-2)=4\).
Solution
Step 1 — Let \(\varepsilon>0\).
Step 2 — Choose \(\delta=\dfrac{\varepsilon}{3}\). From \(|(3x-2)-4|=|3x-6|=3|x-2|<\varepsilon\), we need \(|x-2|<\dfrac{\varepsilon}{3}\).
Step 3 — Assume \(0<|x-2|<\delta\).
Step 4 — Show:
$$ |(3x-2)-4| = |3x-6| = 3|x-2| < 3\delta = 3\cdot\frac{\varepsilon}{3} = \varepsilon. $$Answer: \(\displaystyle\lim_{x\to 2}(3x-2)=4\). \(\quad\square\)
Examples 2.1 and 2.2 were friendly because the function was linear — \(|x-a|\) factored out cleanly. The next example bumps the difficulty to a nonlinear function. The structure stays the same; only Step 2 (choosing δ) gets harder.
Prove that \(\lim_{x\to 1}(2x+1)=3\).
Solution
Step 1 — Let \(\varepsilon>0\). The definition begins "for every \(\varepsilon>0\)", so we open by declaring an arbitrary positive \(\varepsilon\). Whatever follows must work for any choice of \(\varepsilon\).
Step 2 — Choose \(\delta=\dfrac{\varepsilon}{2}\). The definition says "there exists a \(\delta>0\)" — we have to find \(\delta\). Where does \(\dfrac{\varepsilon}{2}\) come from? Two paths to the same answer:
- Algebra: We want to end up with \(|(2x+1)-3|<\varepsilon\). Simplify: \(|2x-2|=|2||x-1|=2|x-1|<\varepsilon\), so we need \(|x-1|<\dfrac{\varepsilon}{2}\). That's exactly the δ we want. - Geometry: The line \(y=2x+1\) has slope 2. To force \(y\)-values within \(\varepsilon\) of \(3\), \(x\) needs to stay within \(\dfrac{\varepsilon}{2}\) of \(1\). See Figure 2.40.
Figure 2.40 — This graph shows how we find \(\delta\) geometrically.
Step 3 — Assume \(0<|x-1|<\delta\). That's the "if" half of the conditional. We assume it and chase the "then" half.
Step 4 — Show \(|(2x+1)-3|<\varepsilon\). Compute:
$$ \begin{aligned} |(2x+1)-3| &= |2x-2| && \text{absolute-value algebra} \\ &= |2||x-1| && \text{factor out } |2|=2 \\ &= 2|x-1| \\ &< 2\cdot\delta && \text{because } |x-1|<\delta \\ &= 2\cdot\frac{\varepsilon}{2}=\varepsilon. && \text{because we chose } \delta=\varepsilon/2. \end{aligned} $$Answer: For every \(\varepsilon>0\), choosing \(\delta=\varepsilon/2\) makes \(|(2x+1)-3|<\varepsilon\) whenever \(0<|x-1|<\delta\). Therefore \(\displaystyle\lim_{x\to 1}(2x+1)=3\). \(\quad\square\)
The next strategy box collects the recipe. Every ε–δ proof in this section follows the same four-step shape.
Problem-Solving Strategy: Proving That \(\lim_{x\to a}f(x)=L\) for a Specific Function
1. Let \(\varepsilon>0\). Open with an arbitrary positive epsilon.
2. Choose \(\delta\) (in terms of \(\varepsilon\)). Work backward from the target inequality \(|f(x)-L|<\varepsilon\) to figure out which δ makes the implication work. This is the only creative step.
3. Assume \(0<|x-a|<\delta\). State the "if" half of the conditional.
4. Show \(|f(x)-L|<\varepsilon\). Use the assumption from Step 3 and the choice from Step 2 to chase the inequality, line by line, down to \(<\varepsilon\). Close with \(\square\).
The next example uses the strategy in a fill-in-the-blank format — it's the same recipe, broken into pieces you can finish.
Find a δ corresponding to a given \(\varepsilon>0\) for a proof that \(\lim_{x\to 9}\sqrt{x}=3\).
Solution
Step 1 — Set up the target. Want \(|\sqrt{x}-3|<\varepsilon\), i.e. \(3-\varepsilon<\sqrt{x}<3+\varepsilon\). Assume \(\varepsilon\le 3\) so \(3-\varepsilon\ge 0\) and we can square.
Step 2 — Square the target. \((3-\varepsilon)^2<x<(3+\varepsilon)^2\), i.e. \(9-6\varepsilon+\varepsilon^2<x<9+6\varepsilon+\varepsilon^2\). Subtract 9: \(-6\varepsilon+\varepsilon^2<x-9<6\varepsilon+\varepsilon^2\).
Step 3 — Pick the smaller distance from 9. The left distance is \(6\varepsilon-\varepsilon^2\); the right distance is \(6\varepsilon+\varepsilon^2\). The left side is smaller, so choose \(\delta=6\varepsilon-\varepsilon^2\) (or any positive value at most this). For a clean small-ε answer, \(\delta=6\varepsilon-\varepsilon^2\) works for all \(0<\varepsilon\le 3\).
Answer: \(\delta=6\varepsilon-\varepsilon^2\) (assuming \(\varepsilon\le 3\)).
The geometric approach above is illuminating but can get clumsy. Many ε–δ proofs are easier with pure algebra — and algebra is the workhorse for the limit-law proofs we'll get to in the next subsection.
Complete the proof that \(\lim_{x\to-1}(4x+1)=-3\) by filling in the blanks.
Solution
Step 1 — Let \(\varepsilon>0\). (Required opening line.)
Step 2 — Choose \(\delta=\dfrac{\varepsilon}{4}\). Reasoning: from the target \(|(4x+1)-(-3)|=|4x+4|=|4||x+1|=4|x+1|<\varepsilon\), we need \(|x+1|<\dfrac{\varepsilon}{4}\).
Step 3 — Assume \(0<|x-(-1)|<\delta\), equivalently \(0<|x+1|<\delta\).
Step 4 — Show:
$$ |(4x+1)-(-3)| = |4x+4| = |4||x+1| < 4\delta = 4\cdot\frac{\varepsilon}{4} = \varepsilon. $$Answer: \(\displaystyle\lim_{x\to-1}(4x+1)=-3\). \(\quad\square\)
Complete the proof that \(\lim_{x\to 1}x^2=1\). Let \(\varepsilon>0\); choose \(\delta=\min\{1,\varepsilon/3\}\); assume \(0<|x-1|<\delta\). Show \(|x^2-1|<\varepsilon\).
Solution
Step 1 — Set up. With \(\delta=\min\{1,\varepsilon/3\}\) and \(0<|x-1|<\delta\), we get \(|x-1|<1\), so \(-1<x-1<1\), so \(0<x<2\), so \(1<x+1<3\). Hence \(|x+1|<3\).
Step 2 — Factor.
$$ |x^2-1| = |x-1|\cdot|x+1| < \delta\cdot 3 \le \frac{\varepsilon}{3}\cdot 3 = \varepsilon. $$Answer: \(\displaystyle\lim_{x\to 1}x^2=1\). \(\quad\square\)
The more complex a function, the more useful the algebraic approach. Now we'll use it to prove a limit law.
Prove that \(\lim_{x\to 2}x^2=4\).
Solution
Step 1 — Let \(\varepsilon>0\). Without loss of generality, assume \(\varepsilon\le 4\). (Why? In Step 2 we'll see that δ involves \(\sqrt{4-\varepsilon}\), so we need \(\varepsilon\le 4\). The clause "without loss of generality" is fine here: if a δ works for ε ≤ 4, the same δ also works for any larger ε.)
Step 2 — Choose \(\delta=\min\!\bigl\{2-\sqrt{4-\varepsilon},\;\sqrt{4+\varepsilon}-2\bigr\}\). Figure 2.41 shows the geometry: the function \(y=x^2\) has different slopes on each side of \(x=2\), so the δ we can use on the left is different from the one on the right. We pick the smaller so it works on both sides.
Figure 2.41 — This graph shows how we find \(\delta\) geometrically for a given \(\varepsilon\) for the proof in Example 2.5.3.
Step 3 — Assume \(0<|x-2|<\delta\). Since the \(0<\) clause just excludes \(x=2\), it's safe to drop and work with \(|x-2|<\delta\), i.e.
$$ -\delta<x-2<\delta. $$Step 4 — Show \(|x^2-4|<\varepsilon\). Because \(\delta\le 2-\sqrt{4-\varepsilon}\), we have \(-(2-\sqrt{4-\varepsilon})\le -\delta\), and because \(\delta\le \sqrt{4+\varepsilon}-2\), we have \(\delta\le\sqrt{4+\varepsilon}-2\). Stitch these to the inequality from Step 3:
$$ -(2-\sqrt{4-\varepsilon})\;\le\;-\delta\;<\;x-2\;<\;\delta\;\le\;\sqrt{4+\varepsilon}-2. $$Simplify the left side and add 2 throughout:
$$ \sqrt{4-\varepsilon}<x<\sqrt{4+\varepsilon}. $$All three parts are positive, so squaring preserves the order:
$$ 4-\varepsilon<x^2<4+\varepsilon. $$Subtract 4 from each part: \(-\varepsilon<x^2-4<\varepsilon\), which is exactly \(|x^2-4|<\varepsilon\).
Answer: \(\displaystyle\lim_{x\to 2}x^2=4\). \(\quad\square\)
Prove that \(\lim_{x\to-1}(x^2-2x+3)=6\).
Solution
Step 1 — Let \(\varepsilon>0\).
Step 2 — Choose \(\delta=\min\{1,\varepsilon/5\}\). Where does this come from? The target inequality factors:
$$ |(x^2-2x+3)-6| = |x^2-2x-3| = |x+1|\cdot|x-3|. $$We want \(|x+1|\cdot|x-3|<\varepsilon\). The tempting choice \(\delta=\varepsilon/|x-3|\) is illegal: δ can't depend on \(x\). The fix: bound \(|x-3|\) by a constant, then divide. If we restrict \(|x+1|<1\), then \(-1<x+1<1\), so \(-2<x<0\), so \(-5<x-3<-3\), giving \(|x-3|<5\). So if \(|x+1|<\delta\le 1\), we have \(|x+1|\cdot|x-3|<5\delta\). To force \(5\delta<\varepsilon\), choose \(\delta\le\varepsilon/5\). Combining: \(\delta=\min\{1,\varepsilon/5\}\).
Step 3 — Assume \(0<|x+1|<\delta\). Then both \(|x+1|<1\) and \(|x+1|<\varepsilon/5\).
Step 4 — Show \(|(x^2-2x+3)-6|<\varepsilon\). From the bound in Step 2, \(|x-3|<5\). Then
$$ |(x^2-2x+3)-6| = |x+1|\cdot|x-3| < \frac{\varepsilon}{5}\cdot 5 = \varepsilon. $$Answer: \(\displaystyle\lim_{x\to-1}(x^2-2x+3)=6\). \(\quad\square\)
2.5.2 Proving Limit Laws
The triangle inequality states that if \(a\) and \(b\) are any real numbers, then
$$ |a+b|\le|a|+|b|. $$The triangle inequality is the engine: it lets us turn a sum of absolute values into a sum we can bound term-by-term. It's named because in two dimensions, \(|a|+|b|\) is the total length of a two-segment trip and \(|a+b|\) is the straight-line shortcut — the shortcut is never longer than the detour.
If \(\lim_{x\to a}f(x)=L\) and \(\lim_{x\to a}g(x)=M\), then \(\lim_{x\to a}\bigl(f(x)+g(x)\bigr)=L+M\).
Proof. Let \(\varepsilon>0\).
Because \(\lim_{x\to a}f(x)=L\), there is a \(\delta_1>0\) such that if \(0<|x-a|<\delta_1\), then \(|f(x)-L|<\varepsilon/2\).
Because \(\lim_{x\to a}g(x)=M\), there is a \(\delta_2>0\) such that if \(0<|x-a|<\delta_2\), then \(|g(x)-M|<\varepsilon/2\).
Choose \(\delta=\min\{\delta_1,\delta_2\}\). Assume \(0<|x-a|<\delta\). Then \(0<|x-a|<\delta_1\) AND \(0<|x-a|<\delta_2\), so both half-ε bounds fire at once.
By the triangle inequality,
$$ \begin{aligned} \bigl|(f(x)+g(x))-(L+M)\bigr| &= \bigl|(f(x)-L) + (g(x)-M)\bigr| \\ &\le |f(x)-L| + |g(x)-M| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{aligned} $$Therefore \(\lim_{x\to a}\bigl(f(x)+g(x)\bigr)=L+M\). \(\quad\square\)
The pattern of this proof — split \(\varepsilon\) in half, find a δ for each piece, take the min, finish with the triangle inequality — recurs in nearly every limit-law proof. It's worth re-reading until the choreography sinks in.
We now flip the question: when does a limit fail to exist? Translate the definition: "\(\lim_{x\to a}f(x)=L\) means for every ε there is a δ such that the implication holds." So \(\lim_{x\to a}f(x)\) does not exist means: for every candidate \(L\), there exists an ε such that for every δ, there's an \(x\) with \(0<|x-a|<\delta\) but \(|f(x)-L|\ge\varepsilon\). The quantifiers all flip.
We now use the ε–δ definition to give a rigorous proof of one of the limit laws from §2.3. We need one tool first.
Show that \(\displaystyle\lim_{x\to 0}\frac{|x|}{x}\) does not exist. The graph of \(f(x)=|x|/x\) is shown below.
Solution
Step 1 — Pick a candidate \(L\) and choose \(\varepsilon=\dfrac{1}{2}\). We'll show no candidate works.
Step 2 — Case 1: \(L\ge 0\). Let \(\delta>0\). Pick \(x=-\delta/2\), so \(x<0\) and \(f(x)=-1\). Then
$$ \left|f(x)-L\right| = |-1-L| = L+1 \ge 1 > \tfrac{1}{2} = \varepsilon. $$So the implication fails: \(0<|x-0|<\delta\) but \(|f(x)-L|\ge\varepsilon\).
Step 3 — Case 2: \(L<0\). Let \(\delta>0\). Pick \(x=\delta/2\), so \(x>0\) and \(f(x)=1\). Then
$$ \left|f(x)-L\right| = |1-L| = 1-L = |L|+1 \ge 1 > \tfrac{1}{2} = \varepsilon. $$Same conclusion: implication fails.
Step 4 — Cover all candidates. Every real \(L\) is either \(\ge 0\) or \(<0\), and we've shown the implication fails in both cases. So no \(L\) qualifies as a limit.
Answer: \(\displaystyle\lim_{x\to 0}\frac{|x|}{x}\) does not exist. \(\quad\square\)
2.5.3 One-Sided and Infinite Limits
Limit from the right. Let \(f(x)\) be defined over an open interval of the form \((a,b)\). Then
$$ \lim_{x\to a^+}f(x)=L $$if for every \(\varepsilon>0\) there exists a \(\delta>0\) such that if \(0<x-a<\delta\), then \(|f(x)-L|<\varepsilon\).
Limit from the left. Let \(f(x)\) be defined over an open interval of the form \((b,a)\) where \(b<a\). Then
$$ \lim_{x\to a^-}f(x)=L $$if for every \(\varepsilon>0\) there exists a \(\delta>0\) such that if \(-\delta<x-a<0\), then \(|f(x)-L|<\varepsilon\).
Let \(f(x)\) be defined for all \(x\ne a\) in an open interval containing \(a\). Then:
- \(\displaystyle\lim_{x\to a}f(x)=+\infty\) if for every \(M>0\) there exists \(\delta>0\) such that if \(0<|x-a|<\delta\), then \(f(x)>M\).
- \(\displaystyle\lim_{x\to a}f(x)=-\infty\) if for every \(M>0\) there exists \(\delta>0\) such that if \(0<|x-a|<\delta\), then \(f(x)<-M\).
Just as we generalized the intuitive limit to one-sided and infinite forms in §2.2, we now generalize the formal definition. The structure is the same — quantifiers, δ in terms of ε — with one tweak: the inequality \(0<|x-a|<\delta\) gets replaced by either \(0<x-a<\delta\) (right) or \(-\delta<x-a<0\) (left), so we only consider \(x\) on the matching side of \(a\).
Find a δ corresponding to a given \(\varepsilon\) for a proof that \(\lim_{x\to 1^-}\sqrt{1-x}=0\).
Solution
Step 1 — Target. Want \(|\sqrt{1-x}-0|<\varepsilon\), i.e. \(0\le\sqrt{1-x}<\varepsilon\), which squares to \(0\le 1-x<\varepsilon^2\), i.e. \(-\varepsilon^2<x-1\le 0\).
Step 2 — Choose δ. \(\delta=\varepsilon^2\). With \(-\delta<x-1<0\), we get \(0<1-x<\delta=\varepsilon^2\), so \(\sqrt{1-x}<\varepsilon\).
Answer: \(\delta=\varepsilon^2\).
For infinite limits, the target inequality changes: instead of "make \(f(x)\) within ε of L", we want "make \(f(x)\) larger than any chosen \(M\)". So the ε bound becomes an M bound, but δ stays a width on \(x\). Figure 2.43 shows the picture: as \(M\) grows, the δ-band around \(a\) where \(f(x)>M\) shrinks.
Figure 2.43 — These graphs plot values of \(\delta\) for \(M\) to show that \(\lim_{x\to a}f(x)=+\infty\).
Prove that \(\lim_{x\to 4^+}\sqrt{x-4}=0\).
Solution
Step 1 — Let \(\varepsilon>0\).
Step 2 — Choose \(\delta=\varepsilon^2\). Working backward from the target \(|\sqrt{x-4}-0|<\varepsilon\): squaring gives \(x-4<\varepsilon^2\), so \(\delta=\varepsilon^2\) is the natural choice. Figure 2.42 shows the geometry.
Figure 2.42 — This graph shows how we find \(\delta\) for the proof in Example 2.5.6.
Step 3 — Assume \(0<x-4<\delta\), i.e. \(0<x-4<\varepsilon^2\).
Step 4 — Show: taking square roots (all parts non-negative), \(0\le\sqrt{x-4}<\varepsilon\). So \(|\sqrt{x-4}-0|<\varepsilon\).
Answer: \(\displaystyle\lim_{x\to 4^+}\sqrt{x-4}=0\). \(\quad\square\)
Problem Set 2.5
In the following exercises, write the appropriate \(\varepsilon\)–\(\delta\) definition for each of the given statements.
Problem 1. \(\displaystyle\lim_{x\to a}f(x)=N\)
Solution
Step 1 — Identify the template: The precise (ε–δ) definition of a limit translates the intuitive idea "\(f(x)\) gets arbitrarily close to \(N\) as \(x\) approaches \(a\)" into a quantified statement about distances.
We need a tolerance \(\varepsilon\) on the output side and a matching tolerance \(\delta\) on the input side. The order of quantifiers matters: \(\varepsilon\) is chosen first (by the challenger), then \(\delta\) must be found in response.
Step 2 — Substitute the symbols for this limit: The function is \(f\), the input variable is \(x\), the point of approach is \(a\), and the proposed limit value is \(N\). Plug these into the template directly.
Answer: \(\lim_{x\to a}f(x)=N\) means that for every \(\varepsilon>0\) there exists a \(\delta>0\) such that if \(0<|x-a|<\delta\), then \(|f(x)-N|<\varepsilon\).
Problem 2. \(\displaystyle\lim_{t\to b}g(t)=M\)
Solution
Step 1 — Identify the template: Apply the ε–δ definition of a limit using the symbols supplied. The variable here is \(t\), the function is \(g\), the input is approaching \(b\), and the proposed limit is \(M\).
Step 2 — Write the input and output inequalities: The "input is close to \(b\) but not equal to \(b\)" condition becomes \(0<|t-b|<\delta\); the "output is within \(\varepsilon\) of \(M\)" condition becomes \(|g(t)-M|<\varepsilon\).
Step 3 — Combine with the correct quantifiers: Universal in \(\varepsilon\), existential in \(\delta\), implication from the input bound to the output bound.
Answer: \(\lim_{t\to b}g(t)=M\) means that for every \(\varepsilon>0\) there exists a \(\delta>0\) such that if \(0<|t-b|<\delta\), then \(|g(t)-M|<\varepsilon\).
Problem 3. \(\displaystyle\lim_{x\to c}h(x)=L\)
Solution
Step 1 — Identify the template: Same precise-definition skeleton, now with function \(h\), approach point \(c\), and limit value \(L\).
Step 2 — Substitute carefully: The input distance \(|x-c|\) is bounded by \(\delta\) (with the strict left inequality \(0<\) to exclude \(x=c\) itself), and the output distance \(|h(x)-L|\) must be bounded by \(\varepsilon\).
Answer: \(\lim_{x\to c}h(x)=L\) means that for every \(\varepsilon>0\) there exists a \(\delta>0\) such that if \(0<|x-c|<\delta\), then \(|h(x)-L|<\varepsilon\).
Problem 4. \(\displaystyle\lim_{x\to a}\varphi(x)=A\)
The following graph of the function \(f\) satisfies \(\displaystyle\lim_{x\to 2}f(x)=2\). In the following exercises, determine a value of \(\delta>0\) that satisfies each statement.
Solution
Step 1 — Identify the template: The Greek symbol \(\varphi\) is just the name of the function — the structure of the definition is unchanged. The input variable is \(x\), the approach point is \(a\), and the proposed limit value is \(A\).
Step 2 — Plug in and quantify: Keep \(\varepsilon\) universally quantified and \(\delta\) existentially quantified; demand the implication from the punctured-input inequality to the output inequality.
Answer: \(\lim_{x\to a}\varphi(x)=A\) means that for every \(\varepsilon>0\) there exists a \(\delta>0\) such that if \(0<|x-a|<\delta\), then \(|\varphi(x)-A|<\varepsilon\).
Problem 5. If \(0<|x-2|<\delta\), then \(|f(x)-2|<1\).
Solution
Step 1 — Set up the geometric question: We have \(\varepsilon=1\) and the limit value \(2\) at \(x=2\). We must find a horizontal half-width \(\delta\) so that the graph over \((2-\delta,2+\delta)\) (with \(x=2\) excluded) stays inside the horizontal band \(1<y<3\).
Step 2 — Solve for the band edges on the curve: Read the graph to locate where the curve crosses \(y=1\) and where it crosses \(y=3\). Because the curve is increasing and concave up and passes through \((1,0.5)\), \((2,2)\), \((3,5)\), the level \(y=1\) is crossed slightly past \(x=1\) (near \(x\approx 1.3\)), and \(y=3\) is crossed a bit past \(x=2\) (near \(x\approx 2.3\)).
Step 3 — Take the smaller side: The left half-width is roughly \(2-1.3=0.7\); the right half-width is roughly \(2.3-2=0.3\). Concavity makes the curve rise more steeply on the right, so the right side gives the binding constraint. We must use the smaller of the two so the band condition holds on both sides.
Step 4 — Choose \(\delta\) with a small safety margin: Picking \(\delta=0.3\) just hits the right boundary; nudge it down to be safely inside.
Answer: \(\delta\approx 0.25\). Then \(0<|x-2|<0.25\) implies \(|f(x)-2|<1\).
Problem 6. If \(0<|x-2|<\delta\), then \(|f(x)-2|<0.5\).
The following graph of the function \(f\) satisfies \(\displaystyle\lim_{x\to 3}f(x)=-1\). In the following exercises, determine a value of \(\delta>0\) that satisfies each statement.
Solution
Step 1 — Tighten the band: Now \(\varepsilon=0.5\), so we need \(1.5<y<2.5\). Find the \(x\)-values where the curve crosses these horizontal lines.
Step 2 — Estimate crossings from the graph: Using the same increasing concave-up curve through \((2,2)\) with neighboring points \((1,0.5)\) and \((3,5)\): \(y=1.5\) is crossed near \(x\approx 1.8\), and \(y=2.5\) is crossed near \(x\approx 2.15\) (the curve climbs faster on the right because of concavity).
Step 3 — Take the smaller half-width: Left side: \(2-1.8=0.2\). Right side: \(2.15-2=0.15\). The right side is the binding constraint.
Step 4 — Choose \(\delta\) with a safety margin: Pick \(\delta\) slightly below the binding \(0.15\).
Answer: \(\delta\approx 0.1\). Then \(0<|x-2|<0.1\) implies \(|f(x)-2|<0.5\).
Problem 7. If \(0<|x-3|<\delta\), then \(|f(x)+1|<1\).
Solution
Step 1 — Use the linear formula: The graph is the line \(y=2-x\) with slope \(-1\). The limit at \(x=3\) is \(f(3)=-1\), so \(\varepsilon=1\) gives the output band \(-2<y<0\).
Step 2 — Convert the output bound to an input bound: For a linear function with \(|\text{slope}|=1\), an output change of \(\varepsilon\) corresponds to an input change of \(\varepsilon/|\text{slope}|=\varepsilon\). Explicitly: $$|f(x)+1|=|(2-x)+1|=|3-x|=|x-3|.$$
So \(|f(x)+1|<1\) is exactly \(|x-3|<1\).
Step 3 — Choose \(\delta\): Take \(\delta=1\) (equality works for a line; no safety margin needed since the slope is constant).
Answer: \(\delta=1\). Then \(0<|x-3|<1\) implies \(|f(x)+1|<1\).
Problem 8. If \(0<|x-3|<\delta\), then \(|f(x)+1|<2\).
The following graph of the function \(f\) satisfies \(\displaystyle\lim_{x\to 3}f(x)=2\). In the following exercises, for each value of \(\varepsilon\), find a value of \(\delta>0\) such that the precise definition of limit holds true.
Solution
Step 1 — Recall the algebra from 2.5.7: For the line \(y=2-x\), \(|f(x)+1|=|x-3|\) exactly.
Step 2 — Translate the new tolerance: With \(\varepsilon=2\), the requirement \(|f(x)+1|<2\) is equivalent to \(|x-3|<2\).
Step 3 — Choose \(\delta\): Set \(\delta=2\); the slope is \(-1\) so the band is captured exactly.
Answer: \(\delta=2\). Then \(0<|x-3|<2\) implies \(|f(x)+1|<2\).
Problem 9. \(\varepsilon=1.5\)
Solution
Step 1 — Read the slope from the graph: The line crosses the \(x\)-axis near \((2.25,0)\) and passes through \((3,2)\), so its slope is approximately $$m=\frac{2-0}{3-2.25}=\frac{2}{0.75}\approx 2.67.$$
The limit at \(x=3\) is \(2\).
Step 2 — Convert output tolerance to input tolerance: For a linear function, an output change of \(\varepsilon\) corresponds to an input change of \(\varepsilon/|m|\). With \(\varepsilon=1.5\): $$\frac{\varepsilon}{|m|}=\frac{1.5}{2.67}\approx 0.56.$$
Step 3 — Choose \(\delta\) with a tiny safety margin: Because the slope was read off a graph, pick \(\delta\) slightly below the computed half-width.
Answer: \(\delta\approx 0.5\). Then \(0<|x-3|<0.5\) implies \(|f(x)-2|<1.5\).
Problem 10. \(\varepsilon=3\)
[T] In the following exercises, use a graphing calculator to find a number \(\delta\) such that the statements hold true.
Solution
Step 1 — Reuse the slope: From 2.5.9, the line has slope \(m\approx 2.67\) and the limit at \(x=3\) is \(2\).
Step 2 — Convert the larger tolerance: With \(\varepsilon=3\): $$\frac{\varepsilon}{|m|}=\frac{3}{2.67}\approx 1.12.$$
Step 3 — Choose \(\delta\) with a safety margin: Take \(\delta\) just under the computed bound so the input band sits safely inside the region where \(|f(x)-2|<3\).
Answer: \(\delta\approx 1\). Then \(0<|x-3|<1\) implies \(|f(x)-2|<3\).
Problem 11. \(\bigl|\sin(2x)-\dfrac{1}{2}\bigr|<0.1\), whenever \(\bigl|x-\dfrac{\pi}{12}\bigr|<\delta\).
Solution
Step 1 — Set up the target inequality: We need \(\bigl|\sin(2x) - \tfrac{1}{2}\bigr| < 0.1\), which means \(0.4 < \sin(2x) < 0.6\).
Step 2 — Solve for \(2x\) near \(\pi/6\): Since \(\sin(\pi/6) = 1/2\) and \(\sin\) is increasing on a neighborhood of \(\pi/6\), use \(\arcsin\):
$$2x_{\text{low}} = \arcsin(0.4) \approx 0.41152, \qquad 2x_{\text{high}} = \arcsin(0.6) \approx 0.64350.$$So \(x_{\text{low}} \approx 0.20576\) and \(x_{\text{high}} \approx 0.32175\), with center \(\pi/12 \approx 0.26180\).
Step 3 — Take the smaller half-width: The distances to \(\pi/12\) are about \(0.05604\) on the left and \(0.05995\) on the right. Choose the smaller so the inequality holds on both sides.
Answer: Any \(\delta \le 0.056\) works (e.g., \(\delta = 0.05\)).
Note on the graphing calculator: Graphing \(y_1 = \sin(2x)\) with horizontal lines \(y = 0.4\) and \(y = 0.6\) and using the intersect tool on a window centered at \(x = \pi/12\) reproduces the endpoints \(x \approx 0.2058\) and \(x \approx 0.3218\); reading the smaller gap to \(\pi/12\) confirms \(\delta \approx 0.056\).
Problem 12. \(\bigl|\sqrt{x-4}-2\bigr|<0.1\), whenever \(|x-8|<\delta\).
In the following exercises, use the precise definition of limit to prove the given limits.
Solution
Step 1 — Set up the target inequality: We need \(\bigl|\sqrt{x-4} - 2\bigr| < 0.1\), i.e., \(1.9 < \sqrt{x-4} < 2.1\).
Step 2 — Square through (all quantities positive):
$$3.61 < x - 4 < 4.41 \implies 7.61 < x < 8.41.$$The center is \(x = 8\), with distances \(|8 - 7.61| = 0.39\) on the left and \(|8.41 - 8| = 0.41\) on the right.
Step 3 — Take the smaller half-width: Choose the smaller so the inequality holds on both sides.
Answer: Any \(\delta \le 0.39\) works (e.g., \(\delta = 0.39\)).
Note on the graphing calculator: Plotting \(y_1 = \sqrt{x-4}\) with horizontal lines \(y = 1.9\) and \(y = 2.1\) and using intersect near \(x = 8\) returns \(x \approx 7.61\) and \(x \approx 8.41\); the smaller distance to \(x = 8\) confirms \(\delta \approx 0.39\).
Problem 13. \(\displaystyle\lim_{x\to 2}(5x+8)=18\)
Solution
Step 1 — Let \(\varepsilon > 0\). We want to find \(\delta > 0\) so that \(0 < |x - 2| < \delta\) implies \(|(5x + 8) - 18| < \varepsilon\).
Step 2 — Choose \(\delta\): Simplify the target:
$$|(5x + 8) - 18| = |5x - 10| = 5|x - 2|.$$To force \(5|x - 2| < \varepsilon\) we need \(|x - 2| < \varepsilon/5\). So choose \(\delta = \varepsilon/5\).
Step 3 — Assume \(0 < |x - 2| < \delta\): Then \(|x - 2| < \varepsilon/5\).
Step 4 — Show \(|f(x) - 18| < \varepsilon\):
$$|(5x + 8) - 18| = 5|x - 2| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon.$$Answer: With \(\delta = \varepsilon/5\), the precise definition is satisfied, so \(\lim_{x \to 2}(5x + 8) = 18\).
Problem 14. \(\displaystyle\lim_{x\to 3}\dfrac{x^2-9}{x-3}=6\)
Solution
Step 1 — Let \(\varepsilon > 0\). We want \(\delta > 0\) so that \(0 < |x - 3| < \delta\) implies \(\left|\tfrac{x^2 - 9}{x - 3} - 6\right| < \varepsilon\).
Step 2 — Simplify on the punctured domain: For \(x \neq 3\),
$$\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3.$$So \(\left|\tfrac{x^2 - 9}{x - 3} - 6\right| = |x + 3 - 6| = |x - 3|\).
Step 3 — Choose \(\delta\): To make \(|x - 3| < \varepsilon\), choose \(\delta = \varepsilon\).
Step 4 — Verify: Assume \(0 < |x - 3| < \delta = \varepsilon\). Since \(x \neq 3\), the simplification is valid, and
$$\left|\frac{x^2 - 9}{x - 3} - 6\right| = |x - 3| < \varepsilon.$$Answer: With \(\delta = \varepsilon\), the definition holds, so \(\lim_{x \to 3}\dfrac{x^2 - 9}{x - 3} = 6\).
Problem 15. \(\displaystyle\lim_{x\to 2}\dfrac{2x^2-3x-2}{x-2}=5\)
Solution
Step 1 — Let \(\varepsilon > 0\). We seek \(\delta > 0\) so that \(0 < |x - 2| < \delta\) implies \(\left|\tfrac{2x^2 - 3x - 2}{x - 2} - 5\right| < \varepsilon\).
Step 2 — Simplify on the punctured domain: Factor the numerator:
$$2x^2 - 3x - 2 = (2x + 1)(x - 2).$$For \(x \neq 2\), \(\dfrac{2x^2 - 3x - 2}{x - 2} = 2x + 1\), so
$$\left|\frac{2x^2 - 3x - 2}{x - 2} - 5\right| = |2x + 1 - 5| = |2x - 4| = 2|x - 2|.$$Step 3 — Choose \(\delta\): To force \(2|x - 2| < \varepsilon\) we need \(|x - 2| < \varepsilon/2\). Choose \(\delta = \varepsilon/2\).
Step 4 — Verify: Assume \(0 < |x - 2| < \delta = \varepsilon/2\). Then \(x \neq 2\), so
$$\left|\frac{2x^2 - 3x - 2}{x - 2} - 5\right| = 2|x - 2| < 2 \cdot \frac{\varepsilon}{2} = \varepsilon.$$Answer: With \(\delta = \varepsilon/2\), the definition holds, so \(\lim_{x \to 2}\dfrac{2x^2 - 3x - 2}{x - 2} = 5\).
Problem 16. \(\displaystyle\lim_{x\to 0}x^4=0\)
Solution
Step 1 — Let \(\varepsilon > 0\). We want \(\delta > 0\) so that \(0 < |x - 0| < \delta\) implies \(|x^4 - 0| < \varepsilon\).
Step 2 — Choose \(\delta\): Note \(|x^4 - 0| = |x|^4\). To make \(|x|^4 < \varepsilon\) it suffices to have \(|x| < \varepsilon^{1/4}\). Choose
$$\delta = \varepsilon^{1/4}.$$Step 3 — Assume \(0 < |x| < \delta = \varepsilon^{1/4}\). Squaring twice (squaring preserves order on nonnegatives):
$$|x| < \varepsilon^{1/4} \implies |x|^2 < \varepsilon^{1/2} \implies |x|^4 < \varepsilon.$$Step 4 — Conclude:
$$|x^4 - 0| = |x|^4 < \varepsilon.$$Answer: With \(\delta = \varepsilon^{1/4}\), the precise definition is satisfied, so \(\lim_{x \to 0} x^4 = 0\).
Problem 17. \(\displaystyle\lim_{x\to 2}(x^2+2x)=8\)
In the following exercises, use the precise definition of limit to prove the given one-sided limits.
Solution
Step 1 — Let \(\varepsilon > 0\). We want \(\delta > 0\) so that \(0 < |x - 2| < \delta\) implies \(|(x^2 + 2x) - 8| < \varepsilon\).
Step 2 — Factor the target:
$$x^2 + 2x - 8 = (x - 2)(x + 4), \qquad |(x^2 + 2x) - 8| = |x - 2|\,|x + 4|.$$Step 3 — Bound \(|x + 4|\) with a preliminary \(\delta\): Restrict \(|x - 2| < 1\), so \(1 < x < 3\) and
$$|x + 4| < 7.$$Under this restriction, \(|(x^2 + 2x) - 8| < 7|x - 2|\). To make this \(< \varepsilon\), require \(|x - 2| < \varepsilon/7\).
Step 4 — Choose \(\delta\): Take \(\delta = \min\{1, \varepsilon/7\}\). Assume \(0 < |x - 2| < \delta\). Then \(|x - 2| < 1\) (so \(|x + 4| < 7\)) and \(|x - 2| < \varepsilon/7\), giving
$$|(x^2 + 2x) - 8| = |x - 2|\,|x + 4| < \frac{\varepsilon}{7} \cdot 7 = \varepsilon.$$Answer: With \(\delta = \min\{1, \varepsilon/7\}\), the precise definition is satisfied, so \(\lim_{x \to 2}(x^2 + 2x) = 8\).
Problem 18. \(\displaystyle\lim_{x\to 5^-}\sqrt{5-x}=0\)
Solution
Step 1 — Let \(\varepsilon > 0\). For a left-hand limit we need \(\delta > 0\) so that \(0 < 5 - x < \delta\) (equivalently \(5 - \delta < x < 5\)) implies \(|\sqrt{5 - x} - 0| < \varepsilon\).
Step 2 — Analyze the target: For \(x < 5\), \(5 - x > 0\) so \(\sqrt{5 - x}\) is real and nonnegative, and \(|\sqrt{5 - x} - 0| = \sqrt{5 - x}\). To force \(\sqrt{5 - x} < \varepsilon\) we need (squaring) \(5 - x < \varepsilon^2\).
Step 3 — Choose \(\delta\): Take \(\delta = \varepsilon^2\).
Step 4 — Verify: Assume \(0 < 5 - x < \delta = \varepsilon^2\). Then \(x < 5\), so the square root is defined, and
$$|\sqrt{5 - x} - 0| = \sqrt{5 - x} < \sqrt{\varepsilon^2} = \varepsilon.$$Answer: With \(\delta = \varepsilon^2\), the precise one-sided definition is satisfied, so \(\lim_{x \to 5^-}\sqrt{5 - x} = 0\).
Problem 19. \(\displaystyle\lim_{x\to 0^+}f(x)=-2\), where \(f(x)=\begin{cases}8x-3,&\text{if } x<0 \\ 4x-2,&\text{if } x\ge 0\end{cases}\).
Solution
Step 1 — Let \(\varepsilon > 0\). For the right-hand limit we need \(\delta > 0\) so that \(0 < x - 0 < \delta\) (i.e., \(0 < x < \delta\)) implies \(|f(x) - (-2)| < \varepsilon\).
Step 2 — Identify the relevant branch: For \(x > 0\) (which falls under the \(x \ge 0\) case), \(f(x) = 4x - 2\). Then
$$|f(x) - (-2)| = |4x - 2 + 2| = |4x| = 4|x| = 4x.$$Step 3 — Choose \(\delta\): To force \(4x < \varepsilon\) we need \(x < \varepsilon/4\). Take \(\delta = \varepsilon/4\).
Step 4 — Verify: Assume \(0 < x < \delta = \varepsilon/4\). Then \(f(x) = 4x - 2\) and
$$|f(x) - (-2)| = 4x < 4 \cdot \frac{\varepsilon}{4} = \varepsilon.$$Answer: With \(\delta = \varepsilon/4\), the precise one-sided definition is satisfied, so \(\lim_{x \to 0^+} f(x) = -2\).
Problem 20. \(\displaystyle\lim_{x\to 1^-}f(x)=3\), where \(f(x)=\begin{cases}5x-2,&\text{if } x<1 \\ 7x-1,&\text{if } x\ge 1\end{cases}\).
In the following exercises, use the precise definition of limit to prove the given infinite limits.
Solution
Step 1 — Let \(\varepsilon > 0\). For the left-hand limit we need \(\delta > 0\) so that \(-\delta < x - 1 < 0\) (i.e., \(1 - \delta < x < 1\)) implies \(|f(x) - 3| < \varepsilon\).
Step 2 — Identify the relevant branch: For \(x < 1\), \(f(x) = 5x - 2\). Then
$$|f(x) - 3| = |5x - 2 - 3| = |5x - 5| = 5|x - 1|.$$Step 3 — Choose \(\delta\): To force \(5|x - 1| < \varepsilon\) we need \(|x - 1| < \varepsilon/5\). Take \(\delta = \varepsilon/5\).
Step 4 — Verify: Assume \(-\delta < x - 1 < 0\), so \(0 < 1 - x < \delta = \varepsilon/5\), hence \(|x - 1| < \varepsilon/5\). Since \(x < 1\), \(f(x) = 5x - 2\), giving
$$|f(x) - 3| = 5|x - 1| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon.$$Answer: With \(\delta = \varepsilon/5\), the precise one-sided definition is satisfied, so \(\lim_{x \to 1^-} f(x) = 3\).
Problem 21. \(\displaystyle\lim_{x\to 0}\dfrac{1}{x^2}=\infty\)
Solution
Step 1 — Set up the precise infinite-limit goal: We must show that for every \(M > 0\) there exists \(\delta > 0\) such that whenever \(0 < |x - 0| < \delta\) we have \(\dfrac{1}{x^2} > M\).
Step 2 — Solve the target inequality for \(|x|\): Since \(M > 0\), the inequality \(\dfrac{1}{x^2} > M\) is equivalent (for \(x \neq 0\)) to
$$x^2 < \frac{1}{M} \quad\Longleftrightarrow\quad |x| < \frac{1}{\sqrt{M}}.$$Step 3 — Choose \(\delta\): Let \(\delta = \dfrac{1}{\sqrt{M}}\).
Step 4 — Verify the implication: Suppose \(0 < |x| < \delta = \dfrac{1}{\sqrt{M}}\). Squaring (both sides positive) gives \(x^2 < \dfrac{1}{M}\). Since \(x \neq 0\), inverting reverses the inequality:
$$\frac{1}{x^2} > M.$$Answer: For every \(M > 0\), choosing \(\delta = 1/\sqrt{M}\) shows \(0 < |x| < \delta \Rightarrow 1/x^2 > M\). Hence \(\displaystyle\lim_{x \to 0}\frac{1}{x^2} = \infty\).
Problem 22. \(\displaystyle\lim_{x\to -1}\dfrac{3}{(x+1)^2}=\infty\)
Solution
Step 1 — Set up the precise infinite-limit goal: Given \(M > 0\), we must produce \(\delta > 0\) so that whenever \(0 < |x - (-1)| = |x+1| < \delta\), we have \(\dfrac{3}{(x+1)^2} > M\).
Step 2 — Solve the target inequality: For \(x \neq -1\),
$$\frac{3}{(x+1)^2} > M \quad\Longleftrightarrow\quad (x+1)^2 < \frac{3}{M} \quad\Longleftrightarrow\quad |x+1| < \sqrt{\tfrac{3}{M}}.$$Step 3 — Choose \(\delta\): Let \(\delta = \sqrt{3/M}\).
Step 4 — Verify the implication: Suppose \(0 < |x+1| < \delta\). Squaring gives \((x+1)^2 < 3/M\). Inverting and multiplying by 3 (both sides positive),
$$\frac{3}{(x+1)^2} > M.$$Answer: For every \(M > 0\), \(\delta = \sqrt{3/M}\) works, so \(\displaystyle\lim_{x \to -1}\frac{3}{(x+1)^2} = \infty\).
Problem 23. \(\displaystyle\lim_{x\to 2}-\dfrac{1}{(x-2)^2}=-\infty\)
Solution
Step 1 — Set up the precise negative-infinite-limit goal: Given \(M > 0\), we must find \(\delta > 0\) such that whenever \(0 < |x - 2| < \delta\), we have \(-\dfrac{1}{(x-2)^2} < -M\).
Step 2 — Solve the target inequality: For \(x \neq 2\),
$$-\frac{1}{(x-2)^2} < -M \;\Longleftrightarrow\; \frac{1}{(x-2)^2} > M \;\Longleftrightarrow\; (x-2)^2 < \frac{1}{M} \;\Longleftrightarrow\; |x-2| < \frac{1}{\sqrt{M}}.$$Step 3 — Choose \(\delta\): Let \(\delta = 1/\sqrt{M}\).
Step 4 — Verify the implication: Suppose \(0 < |x-2| < \delta = 1/\sqrt{M}\). Squaring gives \((x-2)^2 < 1/M\), so \(1/(x-2)^2 > M\), and therefore
$$-\frac{1}{(x-2)^2} < -M.$$Answer: \(\delta = 1/\sqrt{M}\) verifies the precise definition, so \(\displaystyle\lim_{x \to 2}-\frac{1}{(x-2)^2} = -\infty\).
Problem 24. An engineer is using a machine to cut a flat square of Aerogel of area \(144\ \text{cm}^2\). If there is a maximum error tolerance in the area of \(8\ \text{cm}^2\), how accurately must the engineer cut on the side, assuming all sides have the same length? How do these numbers relate to \(\delta\), \(\varepsilon\), \(a\), and \(L\)?
Solution
Step 1 — Translate the engineering tolerance into the limit framework: Let \(s\) denote the side length cut. The area is \(A(s) = s^2\). The target area is \(L = 144\) cm², corresponding to the target side \(a = 12\) cm. The maximum tolerance on the area is \(\varepsilon = 8\) cm². We need the largest \(\delta > 0\) such that \(|s - 12| < \delta\) guarantees \(|s^2 - 144| < 8\).
Step 2 — Factor and bound: Write
$$|s^2 - 144| = |s - 12|\cdot|s + 12|.$$For \(s\) close to \(12\), \(|s + 12| \approx 24\). Restricting \(s\) to within \(1\) of \(12\) gives \(|s + 12| \leq 25\), so
$$|s^2 - 144| \leq 25\,|s - 12|.$$Step 3 — Solve for \(\delta\): To force \(25\,|s - 12| < 8\), require \(|s - 12| < 8/25 = 0.32\). A cleaner estimate uses the linearization \(|s^2 - 144| \approx 24|s - 12|\), giving \(|s - 12| \lesssim 8/24 = 1/3\) cm.
Step 4 — Identify each symbol: In the \(\varepsilon\)–\(\delta\) framing of \(\displaystyle\lim_{s \to a} A(s) = L\),
| Symbol | Engineering meaning | Value |
|---|---|---|
| \(a\) | target side length | \(12\) cm |
| \(L\) | target area | \(144\) cm² |
| \(\varepsilon\) | area tolerance | \(8\) cm² |
| \(\delta\) | required side tolerance | \(\approx 1/3\) cm |
Answer: The side must be cut to within roughly \(\delta \approx 1/3\) cm (more conservatively, \(8/25 = 0.32\) cm) of 12 cm. Here \(a = 12\), \(L = 144\), \(\varepsilon = 8\), and \(\delta \approx 1/3\) play exactly the roles in the precise definition of \(\displaystyle\lim_{s \to 12} s^2 = 144\).
Problem 25. Use the precise definition of limit to prove that the following limit does not exist: \(\displaystyle\lim_{x\to 1}\dfrac{|x-1|}{x-1}\).
Solution
Step 1 — Identify the two-sided behavior: Let \(f(x) = |x-1|/(x-1)\). For \(x > 1\), \(|x-1| = x-1\), so \(f(x) = 1\). For \(x < 1\), \(|x-1| = -(x-1)\), so \(f(x) = -1\). The function jumps by 2 across \(x = 1\).
Step 2 — Set up a proof by contradiction: Suppose, for contradiction, that \(\displaystyle\lim_{x \to 1} f(x) = L\) for some real \(L\). Choose \(\varepsilon = 1/2\). By the precise definition, there exists \(\delta > 0\) such that
$$0 < |x - 1| < \delta \;\Longrightarrow\; |f(x) - L| < \tfrac{1}{2}.$$Step 3 — Test from both sides: Pick \(x_1 = 1 + \delta/2 > 1\); then \(f(x_1) = 1\), so \(|1 - L| < 1/2\), giving \(L > 1/2\). Pick \(x_2 = 1 - \delta/2 < 1\); then \(f(x_2) = -1\), so \(|-1 - L| < 1/2\), giving \(L < -1/2\).
Step 4 — Derive the contradiction: We cannot simultaneously have \(L > 1/2\) and \(L < -1/2\). Hence no real \(L\) satisfies the definition.
Answer: \(\displaystyle\lim_{x \to 1}\frac{|x-1|}{x-1}\) does not exist.
Problem 26. Using precise definitions of limits, prove that \(\displaystyle\lim_{x\to 0}f(x)\) does not exist, given that \(f(x)\) is the ceiling function. (Hint: Try any \(\delta<1\).)
Solution
Step 1 — Identify the two-sided behavior of \(\lceil x \rceil\): For \(0 < x \leq 1\), \(\lceil x \rceil = 1\). For \(-1 < x \leq 0\), \(\lceil x \rceil = 0\). So in any deleted neighborhood of 0, \(f\) takes both the values 0 and 1.
Step 2 — Suppose a limit exists and fix \(\varepsilon\): Suppose \(\displaystyle\lim_{x \to 0}\lceil x \rceil = L\). Choose \(\varepsilon = 1/2\). Then some \(\delta > 0\) (we may shrink so that \(\delta < 1\)) makes
$$0 < |x| < \delta \;\Longrightarrow\; |\lceil x \rceil - L| < \tfrac{1}{2}.$$Step 3 — Plug in points from both sides: Take \(x_1 = \delta/2 \in (0, 1)\): \(\lceil x_1 \rceil = 1\), so \(|1 - L| < 1/2\), i.e. \(L > 1/2\). Take \(x_2 = -\delta/2 \in (-1, 0)\): \(\lceil x_2 \rceil = 0\), so \(|0 - L| = |L| < 1/2\), i.e. \(L < 1/2\).
Step 4 — Contradiction: The two conclusions \(L > 1/2\) and \(L < 1/2\) cannot both hold. No real \(L\) satisfies the precise definition.
Answer: \(\displaystyle\lim_{x \to 0}\lceil x \rceil\) does not exist.
Problem 27. Using precise definitions of limits, prove that \(\displaystyle\lim_{x\to 0}f(x)\) does not exist for \(f(x)=\begin{cases}1 & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational}\end{cases}\). (Hint: Think about how you can always choose a rational number \(0<r<\delta\) but \(|f(r)-0|=1\).)
Solution
Step 1 — Recall the density of rationals and irrationals: Every open interval on \(\mathbb{R}\), no matter how small, contains both rational and irrational numbers. So every deleted neighborhood of 0 contains points where \(f(x) = 1\) and points where \(f(x) = 0\).
Step 2 — Suppose a limit exists and fix \(\varepsilon\): Suppose \(\displaystyle\lim_{x \to 0} f(x) = L\). Choose \(\varepsilon = 1/2\). Then some \(\delta > 0\) gives
$$0 < |x| < \delta \;\Longrightarrow\; |f(x) - L| < \tfrac{1}{2}.$$Step 3 — Choose a rational and an irrational in the \(\delta\)-band: By density, pick a rational \(x_r\) with \(0 < |x_r| < \delta\); then \(f(x_r) = 1\), so \(|1 - L| < 1/2\) and \(L > 1/2\). Pick an irrational \(x_i\) with \(0 < |x_i| < \delta\); then \(f(x_i) = 0\), so \(|L| < 1/2\) and \(L < 1/2\).
Step 4 — Contradiction: We cannot have both \(L > 1/2\) and \(L < 1/2\), so no \(L\) works.
Answer: \(\displaystyle\lim_{x \to 0} f(x)\) does not exist.
Problem 28. Using precise definitions of limits, determine \(\displaystyle\lim_{x\to 0}f(x)\) for \(f(x)=\begin{cases}x & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational}\end{cases}\). (Hint: Break into two cases, \(x\) rational and \(x\) irrational.)
Solution
Step 1 — Conjecture the limit: For rational \(x\), \(f(x) = x\), which tends to 0 as \(x \to 0\). For irrational \(x\), \(f(x) = 0\). Both pieces are squeezed by \(|x|\), so we conjecture \(L = 0\).
Step 2 — Set up the precise definition: Given \(\varepsilon > 0\), we must find \(\delta > 0\) such that
$$0 < |x - 0| < \delta \;\Longrightarrow\; |f(x) - 0| < \varepsilon.$$Step 3 — Bound \(|f(x)|\) uniformly: In both cases,
$$|f(x)| = \begin{cases} |x| & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases} \;\;\leq\;\; |x|.$$Step 4 — Choose \(\delta\): Take \(\delta = \varepsilon\). Then \(0 < |x| < \delta = \varepsilon\) implies \(|f(x)| \leq |x| < \varepsilon\).
Answer: \(\displaystyle\lim_{x \to 0} f(x) = 0\).
Problem 29. Using the function from the previous exercise, use the precise definition of limits to show that \(\displaystyle\lim_{x\to a}f(x)\) does not exist for \(a\ne 0\).
Solution
Step 1 — Behavior of \(f\) near a nonzero \(a\): For rational \(x\) near \(a\), \(f(x) = x\), which is close to \(a \neq 0\). For irrational \(x\) near \(a\), \(f(x) = 0\). So in every deleted neighborhood of \(a\), \(f\) takes values close to \(a\) and values exactly 0.
Step 2 — Suppose the limit exists and fix \(\varepsilon\): Suppose \(\displaystyle\lim_{x \to a} f(x) = L\). Choose \(\varepsilon = |a|/4 > 0\). Then some \(\delta > 0\), which we may shrink so that \(\delta < |a|/4\), gives
$$0 < |x - a| < \delta \;\Longrightarrow\; |f(x) - L| < \tfrac{|a|}{4}.$$Step 3 — Test rational and irrational \(x\) in the \(\delta\)-band: By density, pick an irrational \(x_i\) with \(0 < |x_i - a| < \delta\); then \(f(x_i) = 0\), so
$$|L| = |0 - L| < \tfrac{|a|}{4}.$$Pick a rational \(x_r\) with \(0 < |x_r - a| < \delta\); then \(f(x_r) = x_r\), and since \(|x_r - a| < |a|/4\),
$$|x_r - L| < \tfrac{|a|}{4}.$$Step 4 — Derive the contradiction: Combine the two bounds via the triangle inequality:
$$|a| \;=\; |a - x_r + x_r - L + L| \;\leq\; |a - x_r| + |x_r - L| + |L| \;<\; \tfrac{|a|}{4} + \tfrac{|a|}{4} + \tfrac{|a|}{4} \;=\; \tfrac{3|a|}{4}.$$This forces \(|a| < 3|a|/4\), which is impossible for \(a \neq 0\).
Answer: For every \(a \neq 0\), \(\displaystyle\lim_{x \to a} f(x)\) does not exist.
For the following exercises, suppose that \(\displaystyle\lim_{x\to a}f(x)=L\) and \(\displaystyle\lim_{x\to a}g(x)=M\) both exist. Use the precise definition of limits to prove the following limit laws.
Problem 30. \(\displaystyle\lim_{x\to a}(f(x)+g(x))=L+M\)
Problem 31. \(\displaystyle\lim_{x\to a}[cf(x)]=cL\) for any real constant \(c\). (Hint: Consider two cases: \(c=0\) and \(c\ne 0\).)
Problem 32. \(\displaystyle\lim_{x\to a}[f(x)g(x)]=LM\). (Hint: \(|f(x)g(x)-LM|=|f(x)g(x)-f(x)M+f(x)M-LM|\le|f(x)||g(x)-M|+|M||f(x)-L|\).)
Solutions 30–32
Problem 30
Step 1 — Write what we're given and what we need: Given \(\displaystyle\lim_{x \to a} f(x) = L\) and \(\displaystyle\lim_{x \to a} g(x) = M\). Fix \(\varepsilon > 0\). We must find \(\delta > 0\) such that
$$0 < |x - a| < \delta \;\Longrightarrow\; |(f(x) + g(x)) - (L + M)| < \varepsilon.$$Step 2 — Split \(\varepsilon\) and apply each hypothesis: Apply the definition of \(\lim f = L\) with tolerance \(\varepsilon/2\): there exists \(\delta_1 > 0\) such that \(0 < |x - a| < \delta_1\) implies \(|f(x) - L| < \varepsilon/2\). Similarly, there exists \(\delta_2 > 0\) such that \(0 < |x - a| < \delta_2\) implies \(|g(x) - M| < \varepsilon/2\).
Step 3 — Combine via the triangle inequality: Set \(\delta = \min\{\delta_1, \delta_2\}\). For \(0 < |x - a| < \delta\), both bounds hold simultaneously, so
$$|(f(x) + g(x)) - (L + M)| \;=\; |(f(x) - L) + (g(x) - M)| \;\leq\; |f(x) - L| + |g(x) - M| \;<\; \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} \;=\; \varepsilon.$$Answer: Therefore \(\displaystyle\lim_{x \to a}(f(x) + g(x)) = L + M\).
Problem 31
Step 1 — Handle the trivial case \(c = 0\): If \(c = 0\), then \(c f(x) = 0\) and \(c L = 0\) for all \(x\). For any \(\varepsilon > 0\), any \(\delta > 0\) (say \(\delta = 1\)) gives \(|c f(x) - c L| = 0 < \varepsilon\). So \(\displaystyle\lim_{x \to a} c f(x) = 0 = c L\).
Step 2 — Assume \(c \neq 0\) and set up the precise definition: Fix \(\varepsilon > 0\). We need \(\delta > 0\) such that
$$0 < |x - a| < \delta \;\Longrightarrow\; |c f(x) - c L| < \varepsilon.$$Step 3 — Use the hypothesis with tolerance \(\varepsilon/|c|\): Since \(\displaystyle\lim_{x \to a} f(x) = L\) and \(\varepsilon/|c| > 0\), there exists \(\delta > 0\) such that
$$0 < |x - a| < \delta \;\Longrightarrow\; |f(x) - L| < \frac{\varepsilon}{|c|}.$$Step 4 — Multiply through by \(|c|\): For such \(x\),
$$|c f(x) - c L| \;=\; |c|\,|f(x) - L| \;<\; |c|\cdot\frac{\varepsilon}{|c|} \;=\; \varepsilon.$$Answer: In both cases, \(\displaystyle\lim_{x \to a} c f(x) = c L\).
Problem 32
Step 1 — Use the algebraic identity for \(fg - LM\): Add and subtract \(f(x) M\):
$$f(x) g(x) - L M \;=\; f(x)\bigl(g(x) - M\bigr) + M\bigl(f(x) - L\bigr).$$By the triangle inequality,
$$|f(x) g(x) - L M| \;\leq\; |f(x)|\,|g(x) - M| + |M|\,|f(x) - L|. \qquad(\star)$$Step 2 — Bound \(|f(x)|\) near \(a\): Since \(\displaystyle\lim_{x \to a} f(x) = L\), apply the definition with tolerance 1: there exists \(\delta_1 > 0\) such that \(0 < |x - a| < \delta_1\) implies \(|f(x) - L| < 1\), so
$$|f(x)| \;=\; |f(x) - L + L| \;\leq\; |f(x) - L| + |L| \;<\; |L| + 1.$$Step 3 — Control each piece by \(\varepsilon/2\): Fix \(\varepsilon > 0\). Apply the definition of \(\lim g = M\) with tolerance \(\dfrac{\varepsilon}{2(|L| + 1)}\): there exists \(\delta_2 > 0\) such that \(0 < |x - a| < \delta_2\) implies
$$|g(x) - M| \;<\; \frac{\varepsilon}{2(|L| + 1)}.$$Apply the definition of \(\lim f = L\) with tolerance \(\dfrac{\varepsilon}{2(|M| + 1)}\): there exists \(\delta_3 > 0\) such that \(0 < |x - a| < \delta_3\) implies
$$|f(x) - L| \;<\; \frac{\varepsilon}{2(|M| + 1)}.$$Step 4 — Take \(\delta = \min\{\delta_1, \delta_2, \delta_3\}\) and combine in \((\star)\): For \(0 < |x - a| < \delta\) all three bounds hold, so
$$|f(x) g(x) - L M| \;\leq\; |f(x)|\,|g(x) - M| + |M|\,|f(x) - L|$$ $$<\; (|L| + 1)\cdot\frac{\varepsilon}{2(|L| + 1)} + |M|\cdot\frac{\varepsilon}{2(|M| + 1)} \;<\; \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} \;=\; \varepsilon,$$where in the last step we used \(|M| < |M| + 1\).
Answer: Therefore \(\displaystyle\lim_{x \to a} f(x) g(x) = L M\).
Key Terms
epsilon–delta definition of the limit — \(\lim_{x\to a}f(x)=L\) iff for every \(\varepsilon>0\) there exists \(\delta>0\) such that \(0<|x-a|<\delta\) implies \(|f(x)-L|<\varepsilon\).
universal quantifier — the "for every" phrase that opens an ε–δ definition; the proof must hold for an arbitrary ε.
existential quantifier — the "there exists" phrase that follows; the proof produces a specific δ (in terms of ε).
conditional statement — the "if…then" phrase; the implication the proof must verify under the chosen δ.
triangle inequality — \(|a+b|\le |a|+|b|\); the algebraic tool that lets us bound \(|f+g|\) by \(|f|+|g|\) inside ε–δ proofs.