Calculus 1

The epsilon delta definition of a limit

Saying that outputs “get close to” a number is a picture, not a test. What exact statement decides whether a limit is right, and how do you prove one?

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What a limit is: graphs and tablesPrecalculus

Where you are 11 in Limits

Calculus8

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▴ 2 above What a limit is: graphs and tables One sided limits and when a limit exists

The epsilon delta definition of a limit

The limit laws Continuity5 ▾ Show all 11 (4 more)

Why this exists

In What a limit is: graphs and tables a limit was read off a picture or a table: as x approaches a from both sides, the outputs f(x) crowd around one height L, and we write \lim_{x \to a} f(x) = L. That works when the graph is honest, but “crowd around” is not something you can check. A table can mislead, and two people can look at the same picture and disagree about how close is close.

Kham is a machinist. He cuts square metal plates, and a customer wants each plate to have an area of 9 square centimeters. No cut is perfect, so the customer names a tolerance: the area may be off by at most 0.07 square centimeters. Kham’s question is the reverse one. How close to 3 centimeters must he hold the side so that the area lands inside the tolerance? And if the next customer is stricter, can he always tighten his cut enough to satisfy them?

That second question is the whole idea of a limit. Saying \lim_{x \to 3} x^2 = 9 means that whatever tolerance anyone names for the output, some tolerance on the input guarantees it. This lesson turns that sentence into a definition with no pictures in it, which is what lets limits be proved, and later lets the limit laws, continuity and the derivative rest on something solid.

Definition

Distance on the number line is measured with absolute value: |f(x) - L| is how far the output is from L, and |x - a| is how far the input is from a. By Absolute value inequalities, |x - a| < \delta says the same thing as a - \delta < x < a + \delta.

Let f be defined at every x in some open interval around a, except possibly at a itself, and let L be a real number. Then \lim_{x \to a} f(x) = L means:

> For every number \varepsilon > 0 there is a number \delta > 0 such that, for every x with > 0 < |x - a| < \delta, the output satisfies |f(x) - L| < \varepsilon.

In plain words: name any output tolerance \varepsilon, however small. There is then an input tolerance \delta such that every input within \delta of a, other than a itself, has its output within \varepsilon of L.

Three parts of the statement carry weight.

The order of the quantifiers. “For every \varepsilon” comes first and “there is a \delta” second, so \delta is allowed to depend on \varepsilon: a smaller tolerance may need a tighter window. The lesson Nested quantifiers shows why swapping the two changes the meaning.
The condition 0 < |x - a|. It throws out x = a. A limit describes the outputs near a, never the output at a, so f(a) may be missing or may be any number.
The strict inequalities. Both tolerances are strict, and nothing changes if one of them is relaxed to \le, but the version above is the standard one.

Intuition

Picture the statement on a graph. The output tolerance draws a horizontal band of height 2\varepsilon centered on the line y = L. The input tolerance draws a vertical strip of width 2\delta centered on the line x = a. The limit statement says: however thin the band, a strip can be found so narrow that the graph, inside the strip, never leaves the band (the point above x = a excepted).

Figure 1Inside the strip a - \delta < x < a + \delta the curve stays inside the band L - \varepsilon < y < L + \varepsilon. The value at x = a itself plays no part.

Think of it as a challenge and a reply. A challenger names \varepsilon. You answer with \delta. If every allowed x lands in the band, you win that round. The limit is L exactly when you can win every round, and a proof is a rule that produces a winning \delta from any \varepsilon at once. To show the limit is not L, the challenger needs just one \varepsilon that you cannot answer.

A strip narrower than one that works also works, since it holds fewer inputs. So the task is never to find the best \delta, only some \delta that is small enough.

Results

Interval form. \lim_{x \to a} f(x) = L exactly when, for every \varepsilon > 0, there is a \delta > 0 with L - \varepsilon < f(x) < L + \varepsilon whenever a - \delta < x < a + \delta and x \ne a.

Shrinking. If \delta works for \varepsilon, then every positive \delta' < \delta works for \varepsilon, and \delta works for every \varepsilon' > \varepsilon.

Linear functions. For m \ne 0, \lim_{x \to a} (mx + b) = ma + b, and \delta = \dfrac{\varepsilon}{|m|} works. For m = 0 the function is constant and every \delta works.

Not the limit. \lim_{x \to a} f(x) \ne L (which includes the case where no limit exists) exactly when there is an \varepsilon > 0 such that for every \delta > 0 some x with 0 < |x - a| < \delta has |f(x) - L| \ge \varepsilon.

Uniqueness. A function has at most one limit at a.

Why it is true

Linear functions. Let \varepsilon > 0 and take \delta = \dfrac{\varepsilon}{|m|}. If 0 < |x - a| < \delta, then

\begin{aligned} |(mx + b) - (ma + b)| &= |m(x - a)| = |m|\,|x - a| \\ &< |m|\,\delta = \varepsilon. \end{aligned}

Shrinking. If 0 < |x - a| < \delta' and \delta' < \delta, then 0 < |x - a| < \delta, so |f(x) - L| < \varepsilon, which is also less than any \varepsilon' > \varepsilon.

Not the limit. This is the definition negated one quantifier at a time: “for every \varepsilon” becomes “there is an \varepsilon”, “there is a \delta” becomes “for every \delta”, “for every x” becomes “there is an x”, and the conclusion |f(x) - L| < \varepsilon becomes |f(x) - L| \ge \varepsilon.

Uniqueness. Suppose f(x) has limits L and M at a with L \ne M. Take \displaystyle \varepsilon = \frac{|L - M|}{2}. Each limit gives a \delta; let \delta be the smaller of the two and pick any x with 0 < |x - a| < \delta (one exists because f is defined on an interval around a). Then |f(x) - L| < \varepsilon and |f(x) - M| < \varepsilon. By the Triangle inequality for absolute values, |u + v| \le |u| + |v|,

\begin{aligned} |L - M| &\le |L - f(x)| + |f(x) - M| \\ &< 2\varepsilon = |L - M|, \end{aligned}

and a number cannot be less than itself. So L = M.

Worked examples

The lesson is taught in two parts. Examples 1 to 3 belong to the first: reading \delta off a graph and proving limits of linear functions. Examples 4 to 7 belong to the second: limits of nonlinear functions, where \delta is built from two pieces, and proofs that a limit is not a given number.

Example 1

Reading the largest delta from a graph

For \displaystyle f(x) = \sqrt{x}, \displaystyle \lim_{x \to 1} \sqrt{x} = 1. Take \varepsilon = 0.5. Find the largest \delta that works.

Solution

The band is 0.5 < y < 1.5. Because \displaystyle \sqrt{x} is increasing, it enters and leaves the band where \displaystyle \sqrt{x} = 0.5 and \displaystyle \sqrt{x} = 1.5, that is, at x = 0.5^2 = 0.25 and x = 1.5^2 = 2.25. Every x in 0.25 < x < 2.25 has its output in the band, and no x outside that interval does.

Figure 2Example 1. The band around 1 is reached at x = 0.25 and x = 2.25. The nearer of the two, 0.75 to the left of 1, sets the largest \delta, and the shaded strip is 1 \pm 0.75.

The strip must be centered on 1, so it can reach only as far as the nearer end. The distances from 1 are 1 - 0.25 = 0.75 on the left and 2.25 - 1 = 1.25 on the right. Any \delta larger than 0.75 lets in inputs just below 0.25, such as x = 0.21 when \delta = 0.8, and \displaystyle \sqrt{0.21} \approx 0.458 lies below the band. So the largest \delta is 0.75, and any \delta with 0 < \delta \le 0.75 works.

The two distances differ because the graph is curved: it climbs faster on the left of 1 than on the right, so it leaves the band sooner on the left.

Example 2

A proof for a line

Prove that \lim_{x \to 2} (3x - 1) = 5. Then find a \delta that guarantees an output within 0.03 of 5.

Solution

First, scratch work: find how the output distance depends on the input distance.

|(3x - 1) - 5| = |3x - 6| = 3|x - 2|.

The output distance is exactly three times the input distance, so to make it less than \varepsilon it is enough to make \displaystyle |x - 2| < \frac{\varepsilon}{3}.

Now the proof itself. Let \varepsilon > 0 and choose \displaystyle \delta = \frac{\varepsilon}{3}. If 0 < |x - 2| < \delta, then

|(3x - 1) - 5| = 3|x - 2| < 3\delta = \varepsilon.

Since \varepsilon was any positive number, the limit is 5.

For \varepsilon = 0.03 the rule gives \delta = 0.01: inputs strictly between 1.99 and 2.01 give outputs strictly between 4.97 and 5.03. The scratch work found \delta; the proof only has to check that it works.

Example 3

A falling line with a wide window

Prove that \displaystyle \lim_{x \to 4} \left(7 - \frac{x}{2}\right) = 5.

Solution

Scratch work:

\begin{aligned} \left|\left(7 - \frac{x}{2}\right) - 5\right| &= \left|2 - \frac{x}{2}\right| \\ &= \frac{1}{2}|4 - x| = \frac{1}{2}|x - 4|, \end{aligned}

using |4 - x| = |x - 4|. The output distance is half the input distance, so \delta = 2\varepsilon should work.

Proof. Let \varepsilon > 0 and choose \delta = 2\varepsilon. If 0 < |x - 4| < \delta, then

\left|\left(7 - \frac{x}{2}\right) - 5\right| = \frac{1}{2}|x - 4| < \frac{1}{2}\delta = \varepsilon.

Two points stand out. The slope is negative, but the absolute value removes the sign, which is why the linear result uses |m|. And \delta is larger than \varepsilon: a gentle slope turns a wide window of inputs into a narrow spread of outputs. Nothing in the definition asks \delta to be small compared with \varepsilon.

Example 4

Kham’s plate: a limit of a square

Prove that \lim_{x \to 3} x^2 = 9. Then answer Kham’s question: within what distance of 3 centimeters must he cut the side so the area is within 0.07 square centimeters of 9?

Solution

Scratch work. Factor the output distance:

|x^2 - 9| = |x - 3|\,|x + 3|.

The first factor is the one we control. The second factor changes with x, so it cannot just be divided out as the slope was for a line. Keep it from growing by first agreeing never to take \delta larger than 1. Then |x - 3| < 1 gives 2 < x < 4, so 5 < x + 3 < 7 and |x + 3| < 7. Under that agreement

|x^2 - 9| < 7|x - 3|,

which is less than \varepsilon once \displaystyle |x - 3| < \frac{\varepsilon}{7}. Both conditions are needed, so take the smaller number.

Proof. Let \varepsilon > 0 and choose \displaystyle \delta = \min\left(1, \frac{\varepsilon}{7}\right). If 0 < |x - 3| < \delta, then |x - 3| < 1, so |x + 3| < 7; and \displaystyle |x - 3| < \frac{\varepsilon}{7}. Therefore

|x^2 - 9| = |x - 3|\,|x + 3| < \frac{\varepsilon}{7} \cdot 7 = \varepsilon.

For Kham, \varepsilon = 0.07, so \delta = \min(1, 0.01) = 0.01. He must cut within 0.01 centimeters of 3. Check the worst case: 3.01^2 = 9.0601 and 2.99^2 = 8.9401, both within 0.07 of 9.

The rule gives a safe \delta, not the largest one. The area reaches 9.07 at \displaystyle x = \sqrt{9.07} \approx 3.0116 and drops to 8.93 at \displaystyle x = \sqrt{8.93} \approx 2.9883, which is farther from 3, so the largest \delta is about 0.0116. A proof never needs the largest; it needs one that works for every \varepsilon, and the formula does.

The 1 inside the minimum matters for large \varepsilon. Without it, \varepsilon = 70 would give \delta = 10, and inputs near 13 would be allowed, where |x + 3| is near 16, not below 7.

Example 5

A reciprocal, kept away from zero

Prove that \displaystyle \lim_{x \to 2} \frac{1}{x} = \frac{1}{2}.

Solution

Scratch work. Combine over a common denominator:

\left|\frac{1}{x} - \frac{1}{2}\right| = \left|\frac{2 - x}{2x}\right| = \frac{|x - 2|}{2|x|}.

The factor \displaystyle \frac{1}{2|x|} grows without bound as x nears 0, so it must be kept in check. Agree that \delta \le 1. Then |x - 2| < 1 gives 1 < x < 3, so 2|x| > 2 and \displaystyle \frac{1}{2|x|} < \frac{1}{2}. Under that agreement

\left|\frac{1}{x} - \frac{1}{2}\right| < \frac{1}{2}|x - 2|,

which is less than \varepsilon once |x - 2| < 2\varepsilon.

Proof. Let \varepsilon > 0 and choose \delta = \min(1, 2\varepsilon). If 0 < |x - 2| < \delta, then x > 1, so \displaystyle \frac{1}{2|x|} < \frac{1}{2}, and |x - 2| < 2\varepsilon. Therefore

\begin{aligned} \left|\frac{1}{x} - \frac{1}{2}\right| &= \frac{|x - 2|}{2|x|} \\ &< \frac{1}{2} \cdot 2\varepsilon = \varepsilon. \end{aligned}

The first agreement could have been \displaystyle \frac{1}{2} instead of 1, with a different constant in the second part. It could not have been 2: the strip 0 < x < 4 comes as close to 0 as you like, and no bound on \displaystyle \frac{1}{2|x|} holds there.

Example 6

Showing a proposed limit is wrong

Noy says that \lim_{x \to 3} x^2 = 9.5. Use the definition to show that Noy is wrong.

Solution

By the “not the limit” result, we must find one \varepsilon that no \delta can answer: for every \delta > 0 we must produce an x with 0 < |x - 3| < \delta and |x^2 - 9.5| \ge \varepsilon.

Near 3 the squares are near 9, about 0.5 below Noy’s number, so any \varepsilon a little less than 0.5 should succeed. Take \varepsilon = 0.1. Given any \delta > 0, let \displaystyle x = 3 + \frac{1}{2}\min(\delta, 0.1). Then 0 < x - 3 < \delta, so x is an allowed input, and 3 < x \le 3.05. Squaring positive numbers keeps the order, so

\begin{aligned} x^2 &\le 3.05^2 = 9.3025, \\ 9.5 - x^2 &\ge 0.1975 \ge 0.1. \end{aligned}

Every \delta fails for \varepsilon = 0.1, so 9.5 is not the limit. Combined with Example 4 and uniqueness, 9 is the only number that is.

The \min(\delta, 0.1) is there so that one argument covers a huge \delta as well as a tiny one: it keeps the chosen x near 3 whatever \delta is.

Example 7

A jump: no number is the limit

Let H(x) = 1 for x \ge 0 and H(x) = -1 for x < 0. Show that \lim_{x \to 0} H(x) \ne 1, and then that \lim_{x \to 0} H(x) does not exist.

Solution

First, 1 is not the limit. Take \varepsilon = 1. For any \delta > 0, the input \displaystyle x = -\frac{\delta}{2} satisfies 0 < |x| < \delta and H(x) = -1, so |H(x) - 1| = 2 \ge 1. No \delta answers \varepsilon = 1.

Figure 3Example 7. The pieces sit at heights 1 and -1, two units apart, and a band of total height 2 around any L cannot hold both.

Now any number L at all. Again take \varepsilon = 1, and suppose some \delta worked. The inputs \displaystyle \frac{\delta}{2} and \displaystyle -\frac{\delta}{2} both lie within \delta of 0, so both outputs would be within 1 of L: |1 - L| < 1 and |-1 - L| < 1. The triangle inequality then gives

\begin{aligned} 2 = |1 - (-1)| &\le |1 - L| + |L - (-1)| \\ &< 1 + 1 = 2, \end{aligned}

which is impossible. So no L is the limit, and the limit does not exist. In the picture, the band has total height 2, and two outputs exactly 2 apart can never both lie strictly inside it. The same reasoning, with one-sided strips, is behind One sided limits and when a limit exists.

Boundaries

The definition needs f defined on an open interval around a, with a itself allowed to be missing. If the domain reaches a from one side only, as \displaystyle \sqrt{x} does at 0, the two-sided statement does not apply; the one-sided version replaces 0 < |x - a| < \delta by a < x < a + \delta or by a - \delta < x < a.
L must be a real number, and a a real number. Infinite limits and limits as x \to \infty use a different pair of tolerances, a large height or a large input in place of \varepsilon or \delta; see Infinite limits and vertical asymptotes and Limits at infinity and horizontal asymptotes.
The definition checks a proposed L; it does not produce one. The value is found first, from a graph, a table or algebra, and then proved.
\delta depends on \varepsilon and, for nonlinear functions, on a as well. The rule \displaystyle \delta = \min\left(1, \frac{\varepsilon}{7}\right) belongs to x^2 at 3 and fails at other points.
A \delta from a restricting agreement, such as “\delta \le 1”, is valid only with that agreement kept inside the minimum.

Where this goes next

Proving every limit straight from the definition would be slow, and it is not how limits are usually found. Instead the definition is used once, to prove general rules: the limit of a sum is the sum of the limits, and so on. Those are the limit laws, and their proofs are \varepsilon-\delta arguments like the ones above, with the “smaller of two deltas” move doing most of the work. Continuity at a is the same definition with L replaced by f(a), and the squeeze argument in The squeeze theorem and the limit of sine x over x rests on it too. The style of reasoning, a rule that answers every challenge, and a single counterexample that refutes, is one of the standard Proof techniques.

Exercises

Exercises 1 to 10 go with the first part of the lesson (Examples 1 to 3), Exercises 11 to 19 with the second (Examples 4 to 7), and the reasoning group draws on both.

Part 1: delta from a graph or a formula

For \displaystyle \lim_{x \to 9} \sqrt{x} = 3 and \varepsilon = 1, find the interval of inputs whose outputs lie in the band, and the largest \delta that works.
For \lim_{x \to 2} x^2 = 4 and \varepsilon = 1, find the largest \delta that works, exactly and to three decimal places.
For \displaystyle \lim_{x \to 4} \frac{1}{x} = \frac{1}{4} and \displaystyle \varepsilon = \frac{1}{12}, find the largest \delta that works.
For \lim_{x \to 1} (4x + 3) = 7, find the largest \delta that guarantees an output within 0.02 of 7.
Prove that \lim_{x \to -2} (5x + 4) = -6.
Prove that \displaystyle \lim_{x \to 6} \left(2 - \frac{x}{3}\right) = 0.
Prove that \lim_{x \to a} c = c for a constant c.
Kham also cuts square plates whose perimeter 4s must be within 0.2 centimeters of 36 centimeters. Within what distance of 9 centimeters must the side s be?
Souk tries \displaystyle \delta = \frac{\varepsilon}{2} for \lim_{x \to 1} (4x + 3) = 7. With \varepsilon = 0.02, find an allowed input whose output is not within \varepsilon of 7.
For \lim_{x \to 1} (4x + 3) = 7 and \varepsilon = 0.1, which of \delta = 0.01, \delta = 0.025 and \delta = 0.04 work?

Part 2: nonlinear limits and limits that fail

Prove that \lim_{x \to -1} x^2 = 1.
Prove that \lim_{x \to 1} (x^2 + 3x) = 4.
Prove that \displaystyle \lim_{x \to 3} \frac{1}{x} = \frac{1}{3}.
A customer wants plates of area 16 square centimeters, give or take 0.09. Using the method of Example 4, find a tolerance on the side that guarantees this.
Prove that \lim_{x \to 2} x^3 = 8.
Noy drops the 1 and uses \displaystyle \delta = \frac{\varepsilon}{7} for \lim_{x \to 3} x^2 = 9. With \varepsilon = 70, find an allowed input whose output is not within \varepsilon of 9.
Show from the definition that \lim_{x \to 1} (4x + 3) \ne 7.2.
Let G(x) = 2 for x < 1 and G(x) = 5 for x \ge 1. Show that \lim_{x \to 1} G(x) \ne 5, and that \lim_{x \to 1} G(x) does not exist.
Let g(x) = x^2 for x \ne 3 and g(3) = 0. What is \lim_{x \to 3} g(x)? Does the proof in Example 4 need any change?

Reasoning

Explain why the definition asks for 0 < |x - a| < \delta rather than |x - a| < \delta. Which limit in this lesson would become false without the 0 <?
Suppose some single \delta > 0 worked for every \varepsilon > 0 at once. Prove that f(x) = L for every x with 0 < |x - a| < \delta.
Prove that \displaystyle \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0, using |\sin t| \le 1.
Redo the scratch work of Example 4 with the agreement \displaystyle \delta \le \frac{1}{2} instead of \delta \le 1. What \delta results?
In Example 5, explain why the first agreement could not be \delta \le 2, and name one other agreement that would work.

Answers

4 < x < 16; \delta = 5. The band is \displaystyle 2 < \sqrt{x} < 4, so 4 < x < 16; the distances from 9 are 5 and 7, and the smaller one is the largest \delta.
\displaystyle \delta = \sqrt{5} - 2 \approx 0.236. Positive inputs with 3 < x^2 < 5 are \displaystyle \sqrt{3} < x < \sqrt{5}; the distances from 2 are \displaystyle 2 - \sqrt{3} \approx 0.268 and \displaystyle \sqrt{5} - 2 \approx 0.236.
\delta = 1. The band \displaystyle \frac{1}{6} < \frac{1}{x} < \frac{1}{3} means 3 < x < 6; the distances from 4 are 1 and 2.
\delta = 0.005. |(4x + 3) - 7| = 4|x - 1| < 0.02 exactly when |x - 1| < 0.005.
\displaystyle \delta = \frac{\varepsilon}{5}. |(5x + 4) - (-6)| = |5x + 10| = 5|x + 2| < 5\delta = \varepsilon whenever 0 < |x + 2| < \delta.
\delta = 3\varepsilon. \displaystyle \left|2 - \frac{x}{3}\right| = \frac{1}{3}|x - 6| < \frac{1}{3}\delta = \varepsilon whenever 0 < |x - 6| < \delta.
Any \delta works, say \delta = 1: |c - c| = 0 < \varepsilon for every x.
Within 0.05 centimeters. |4s - 36| = 4|s - 9| < 0.2 exactly when |s - 9| < 0.05.
For example x = 1.009: then \delta = 0.01 allows it, but 4(1.009) + 3 = 7.036, which is 0.036 from 7, more than 0.02.
\delta = 0.01 and \delta = 0.025 work; \delta = 0.04 does not. The largest is \displaystyle \frac{0.1}{4} = 0.025; with \delta = 0.04 the input x = 1.03 gives 7.12.
\displaystyle \delta = \min\left(1, \frac{\varepsilon}{3}\right). |x^2 - 1| = |x + 1|\,|x - 1|; with |x + 1| < 1, -2 < x < 0 and |x - 1| < 3, so \displaystyle |x^2 - 1| < \frac{\varepsilon}{3} \cdot 3 = \varepsilon.
\displaystyle \delta = \min\left(1, \frac{\varepsilon}{6}\right). |x^2 + 3x - 4| = |x - 1|\,|x + 4|; with |x - 1| < 1, 0 < x < 2 and |x + 4| < 6.
\delta = \min(1, 6\varepsilon). \displaystyle \left|\frac{1}{x} - \frac{1}{3}\right| = \frac{|x - 3|}{3|x|}; with |x - 3| < 1, x > 2 and \displaystyle \frac{1}{3|x|} < \frac{1}{6}, so the distance is less than \displaystyle \frac{1}{6} \cdot 6\varepsilon = \varepsilon.
Within 0.01 centimeters of 4. |x^2 - 16| = |x - 4|\,|x + 4|; with |x - 4| < 1, |x + 4| < 9, so \displaystyle \delta = \min\left(1, \frac{\varepsilon}{9}\right) = \min(1, 0.01) = 0.01. Check: 4.01^2 = 16.0801.
\displaystyle \delta = \min\left(1, \frac{\varepsilon}{19}\right). |x^3 - 8| = |x - 2|\,|x^2 + 2x + 4|; with |x - 2| < 1, 1 < x < 3 and x^2 + 2x + 4 < 9 + 6 + 4 = 19.
For example x = 12: \delta = 10 allows it, but 12^2 - 9 = 135, more than 70.
Take \varepsilon = 0.1. For any \delta, let \displaystyle x = 1 + \frac{1}{2}\min(\delta, 0.01). Then 0 < x - 1 < \delta and 7 < 4x + 3 \le 7.02, so |(4x + 3) - 7.2| \ge 0.18 \ge 0.1.
Not 5: take \varepsilon = 1; for any \delta, \displaystyle x = 1 - \frac{\delta}{2} gives G(x) = 2 and |2 - 5| = 3 \ge 1. No limit: take \varepsilon = 1.5; the inputs \displaystyle 1 \pm \frac{\delta}{2} give 2 and 5, and 3 = |5 - 2| \le |5 - L| + |L - 2| < 1.5 + 1.5 = 3 is impossible.
The limit is 9, and the proof needs no change. The definition only looks at 0 < |x - 3| < \delta, so the value g(3) is never used.
With |x - a| < \delta the input a itself would count, so f(a) would have to be within every \varepsilon of L, forcing f(a) = L. Then \lim_{x \to 3} g(x) = 9 in Exercise 19 would be false, and a limit could not exist where f(a) is undefined.
Fix x with 0 < |x - a| < \delta. Then |f(x) - L| < \varepsilon for every \varepsilon > 0. A nonnegative number below every positive number is 0, so f(x) = L.
\delta = \varepsilon. For 0 < |x| < \delta, the bound |\sin t| \le 1 with \displaystyle t = \frac{1}{x} gives \displaystyle \left|x \sin\left(\frac{1}{x}\right)\right| \le |x|, and |x| < \delta = \varepsilon.
\displaystyle \delta = \min\left(\frac{1}{2}, \frac{2\varepsilon}{13}\right). With \displaystyle |x - 3| < \frac{1}{2}, 5.5 < x + 3 < 6.5, so |x^2 - 9| < 6.5|x - 3|, which is below \varepsilon once \displaystyle |x - 3| < \frac{\varepsilon}{6.5} = \frac{2\varepsilon}{13}.
With |x - 2| < 2 the inputs fill 0 < x < 4, which comes as close to 0 as you like, so \displaystyle \frac{1}{2|x|} has no upper bound there. Any agreement below 2 works, for example \displaystyle \delta \le \frac{1}{2}: then \displaystyle x > \frac{3}{2}, \displaystyle \frac{1}{2|x|} < \frac{1}{3}, and \displaystyle \delta = \min\left(\frac{1}{2}, 3\varepsilon\right).