The Limit of a Function - Maths: an open, verified university mathematics reference

Why this matters¶

You drop a stone from a bridge. How fast is it falling exactly one second later? A speedometer would need to measure a speed at a single instant, but speed is distance divided by time, and in a single instant the stone covers no distance in no time.

What we can compute are average speeds. Near the Earth’s surface the stone has fallen about $s(t) = 5t^2$ metres after $t$ seconds. Over the time interval from 1 to $1 + h$ seconds it falls $s(1 + h) - s(1)$ metres, so its average speed is

\frac{s(1 + h) - s(1)}{h} = \frac{5(1+h)^2 - 5}{h} \quad\text{metres per second.}

(1)

(For $h < 0$ this is the average over the interval from $1 + h$ to 1.) We cannot put $h = 0$ : that gives $\frac{0}{0}$ . But we can make $h$ small.

Figure 1:Graph of the average speed $\frac{5(1+h)^2 - 5}{h}$ for $-1 \le h \le 1$ , with a hole at $h = 0$ , where the formula is undefined, and a point on the graph that a slider for $h$ moves. A table lists the values for $h = \pm 0.1, \pm 0.01, \pm 0.001$ : 10.5, 10.05, 10.005 and 9.995, 9.95, 9.5. They approach 10 from both sides.

Try this: move $h$ towards 0 from both sides. Which number do the average speeds approach? What happens when you try to put $h = 0$ in the formula?

The average speeds approach 10, so the stone’s speed at $t = 1$ should be 10 metres per second. This page makes “approach” precise. That matters, because tables and pictures can mislead. The sine function appears in the next figure and the paragraph after it, the worked example on $\sin(1/x)$ , the common mistake “Checking one $\eps$ , or a few points”, and the exercise on $\sin(\pi/x)$ (its statement, hints and solution). They use only the facts about it in the box below.

Figure 2:Graph of $y = \sin(\pi / x)$ for $-1 \le x \le 1$ , $x \ne 0$ . It stays between -1 and 1. Near $x = 0$ it rises to height 1 and falls back through 0 again and again, faster and faster, too fast to draw: the plotted curve joins finitely many computed points, so close to 0 it is a jumble of lines that misses most of the swings, with even a short flat piece at height 0, an artefact of the drawing. A table lists the values at $x = 0.1, 0.01, 0.001$ , where the exact values are $\sin(10\pi) = \sin(100\pi) = \sin(1000\pi) = 0$ (the computer shows tiny numbers such as $-1.2 \times 10^{-15}$ instead, because it rounds $\pi$ ), and at $x = 0.08, 0.016, 0.0032$ , where the values are $\sin(12.5\pi) = \sin(62.5\pi) = \sin(312.5\pi) = 1$ .

Try this: cover the last three rows of the table. What limit at 0 do the first three suggest? Now look at the last three. Does the graph near 0 settle near one height?

Both halves of the table are correct, and they suggest different answers. A table checks finitely many points, and a graph is drawn through finitely many points, so neither can tell us what happens at every point near 0. We need a definition that speaks about all of them. With it, we will show that the average speeds really do approach 10 (Example 2) and that $\sin(\pi/x)$ approaches no number at all (Exercise 9).

What a limit is¶

Informally, $\lim_{x \to a} f(x) = L$ means that $f(x)$ is as close to $L$ as we like for all $x$ close enough to $a$ , but not equal to $a$ . “As close as we like” is a tolerance, which we call $\eps$ , and “close enough” is a distance, which we call $\delta$ .

We measure closeness with distances: $\abs{f(x) - L}$ is the distance from $f(x)$ to $L$ , and $\abs{x - a}$ the distance from $x$ to $a$ . The set of $x$ with $0 < \abs{x - a} < \delta$ is the open interval $(a - \delta, a + \delta)$ with its centre $a$ removed, the punctured neighbourhood of $a$ of radius $\delta$ , by part (e) of the proposition on distance inequalities.

In words. Whatever tolerance $\eps$ someone demands around $L$ , you can answer with a window of half-width $\delta$ around $a$ such that every $x$ in that window, other than $a$ itself, has $f(x)$ within the tolerance. Think of it as a game. A challenger names an $\eps > 0$ ; you reply with a $\delta > 0$ ; you win the round if every $x$ in your window passes. The limit is $L$ when you can win every round. Three points to notice:

$\delta$ is chosen after $\eps$ , so it may depend on $\eps$ (and on $a$ ), but it is one number for the whole window: it may not depend on $x$ .
The window leaves out $x = a$ . The value $f(a)$ , and whether it exists, plays no part.
If a $\delta$ wins a round, so does every smaller $\delta > 0$ : a smaller window has fewer points to check. Since $f$ is defined on an open interval around $a$ (except at $a$ ), a small enough window lies inside that interval, so the requirement that $f$ be defined in the window only rules out windows that are too wide.

Example. $\lim_{x \to a} x = a$ for every real $a$ . Given $\eps > 0$ , take $\delta = \eps$ . If $0 < \abs{x - a} < \delta$ , then $\abs{f(x) - a} = \abs{x - a} < \eps$ .

Non-example. $\lim_{x \to 1} x$ is not 1.5. Take $\eps = 0.25$ . Whatever $\delta > 0$ is offered, the window contains the point $x = 1 + d$ with $d = \frac12 \min(\delta, 0.25)$ , and for it $\abs{f(x) - 1.5} = \abs{d - 0.5} = 0.5 - d \ge 0.375$ , which is not less than 0.25. So no $\delta$ wins the round $\eps = 0.25$ , and 1.5 is not the limit. One round that cannot be won is enough.

The game is easiest to understand by playing it. In the figure, the claim is $\lim_{x \to 2} x^2 = 4$ .

Figure 3:Graph of $y = x^2$ for $1 \le x \le 3$ , with a horizontal band of half-width $\eps$ around $y = 4$ (translucent, dashed edges) and a vertical window $0 < \abs{x - 2} < \delta$ (translucent, solid edges; the dotted line $x = 2$ itself is left out). A slider sets $\eps$ from 0.05 to 1.5 in steps of 0.05, starting at 0.5. For each $\eps$ the widget states the largest $\delta$ it found on each side of 2, and how many points it checked, and then the smaller of the two as the largest $\delta$ it found for both sides. For $\eps = 0.5$ it reports 0.129171 on the left and 0.12132 on the right; the exact values are $2 - \sqrt{3.5} \approx 0.129171$ and $\sqrt{4.5} - 2 \approx 0.121320$ . A second slider sets your own $\delta$ , from 0.001 to 1 in steps of 0.001, and a button sets it to the largest value on that slider that is at most the $\delta$ the widget found. Points of the graph inside the window but outside the band are marked with crosses, and a sentence says whether your $\delta$ passed the widget’s check at the points it tested, or names a point where it fails.

Try this: play the ε–δ game.

Select the $\eps$ slider and press ← eight times, to $\eps = 0.1$ . Read the largest $\delta$ the widget found on the left of 2 and on the right. (It reports 0.0251582 on the left and 0.0248456 on the right.)
You must answer with one $\delta$ . Which of the two numbers can you use, and why not the other?
Press Set δ to the largest on the slider ( $\delta = 0.024$ ), then select the $\delta$ slider and press → once ( $\delta = 0.025$ ). Where does the widget see your $\delta$ fail?
Try $\eps = 0.05$ and $\eps = 1$ . What happens to the $\delta$ the widget finds as $\eps$ shrinks?

Why the smaller one. The window $0 < \abs{x - 2} < \delta$ reaches the same distance $\delta$ on both sides of 2, so one $\delta$ has to work on the left and on the right at once. For $\eps = 0.1$ , a $\delta$ larger than $\sqrt{4.1} - 2 \approx 0.024846$ (the widget found 0.0248456) lets in points just to the right of $\sqrt{4.1}$ , where $x^2$ is already outside the band (at $x = 2.025$ , $x^2 = 4.100625$ ); the left side would tolerate up to $2 - \sqrt{3.9} \approx 0.025158$ , but that does not help. So the $\delta$ for both sides is the smaller of the two, here the right side’s, because $x^2$ grows faster to the right of 2 than it falls to the left. As $\eps$ shrinks, the band narrows and so does the window.

The widget’s numbers are observations: it checks finitely many points, so it cannot prove that its $\delta$ works for every $x$ in the window, or that a $\delta$ exists for every $\eps$ . Exercise Exercise 5 finds the exact largest $\delta$ for $\eps = 0.1$ , and the rigorous track proves, in Example 6, that $\delta = \min(1, \eps/5)$ works for every $\eps > 0$ . That $\delta$ is smaller than the largest one (for $\eps = 0.1$ it is 0.02): a proof only needs some $\delta$ that works.

Remark 1 (The value at

a

does not matter)

The definition only looks at $x$ with $0 < \abs{x - a}$ , that is, at $x \ne a$ . So $\lim_{x \to a} f(x)$ depends only on the values of $f$ near $a$ : changing $f(a)$ , or leaving $f(a)$ undefined, does not change the limit or whether it exists.

Example. $g(x) = \dfrac{x^2 - 1}{x - 1}$ is not defined at $x = 1$ , but $\lim_{x \to 1} g(x) = 2$ . For $x \ne 1$ we may cancel $x - 1 \ne 0$ : $g(x) = \dfrac{(x - 1)(x + 1)}{x - 1} = x + 1$ , so $\abs{g(x) - 2} = \abs{x - 1}$ , and $\delta = \eps$ works.

Non-example. The limit need not be the value at $a$ . Let $h(x) = x + 1$ for $x \ne 1$ and $h(1) = 5$ . For $x \ne 1$ , $h(x) = g(x)$ , and only those $x$ enter the definition, so again $\delta = \eps$ works and $\lim_{x \to 1} h(x) = 2$ , although $h(1) = 5$ . Putting $x = 1$ into $h$ gives the wrong number.

Main results¶

Could a function have two different limits at the same point? The game suggests not: if $f(x)$ is within any tolerance of $L$ for all $x$ close enough to $a$ , it cannot also be within every tolerance of a different number $M$ , because $L$ and $M$ are a fixed distance apart.

Proof (Rigorous track)

We argue by contradiction: if $L \ne M$ , we take a tolerance of half their distance and find one $x$ whose value $f(x)$ would have to be within that tolerance of both, which the triangle inequality forbids.

Suppose $L \ne M$ , and let $\eps = \frac{1}{2}\abs{L - M}$ , which is positive. Since $\lim_{x \to a} f(x) = L$ , there is a $\delta_1 > 0$ such that $f$ is defined and $\abs{f(x) - L} < \eps$ for every $x$ with $0 < \abs{x - a} < \delta_1$ . Since $\lim_{x \to a} f(x) = M$ , there is a $\delta_2 > 0$ such that $f$ is defined and $\abs{f(x) - M} < \eps$ for every $x$ with $0 < \abs{x - a} < \delta_2$ .

Let $\delta = \min(\delta_1, \delta_2)$ and $x = a + \frac{\delta}{2}$ . Then $0 < \abs{x - a} = \frac{\delta}{2} < \delta$ , so $x$ lies in both windows: $f(x)$ is defined, and $\abs{f(x) - L} < \eps$ and $\abs{f(x) - M} < \eps$ . The first of these says also that $\abs{L - f(x)} < \eps$ , because $\abs{L - f(x)} = \abs{f(x) - L}$ by property 2 of the absolute value. By part (b) of the triangle inequality, $\abs{u - v} \le \abs{u - w} + \abs{w - v}$ , with $u = L$ , $v = M$ and $w = f(x)$ ,

\abs{L - M} \le \abs{L - f(x)} + \abs{f(x) - M} < \eps + \eps = \abs{L - M},

(4)

that is, $\abs{L - M} < \abs{L - M}$ , which is impossible. So $L = M$ .

Because of Theorem 1, we may speak of the limit of $f(x)$ as $x \to a$ , when there is one. If there is none, we say that the limit does not exist. The proof used the requirement in Definition 1 that $f$ be defined at every point of the window; the rigorous track shows that if the definition dropped its requirements on the domain of $f$ , uniqueness would fail.

Worked examples¶

Proving a limit from the definition¶

Each proof has the same shape. We start from $\abs{f(x) - L}$ and rewrite it in terms of $\abs{x - a}$ (the scratch work); this tells us which $\delta$ to choose; then we check that this $\delta$ works.

Example 1 (Prove

\lim_{x \to 3} (2x - 1) = 5

from the definition)

Goal. Given $\eps > 0$ , find $\delta > 0$ such that $0 < \abs{x - 3} < \delta$ implies $\abs{(2x - 1) - 5} < \eps$ . The function $2x - 1$ is defined for every $x$ , so every window lies in its domain.

Scratch work. $\abs{(2x - 1) - 5} = \abs{2x - 6} = 2\abs{x - 3}$ .
Choose $\delta$ . We need $2\abs{x - 3} < \eps$ , that is, $\abs{x - 3} < \frac{\eps}{2}$ . So take $\delta = \frac{\eps}{2}$ . It depends on $\eps$ only, not on $x$ .
Proof. Let $\eps > 0$ and $\delta = \frac{\eps}{2}$ . If $0 < \abs{x - 3} < \delta$ , then
$\abs{(2x - 1) - 5} = 2\abs{x - 3} < 2 \cdot \frac{\eps}{2} = \eps .$
(5)
So for every $\eps > 0$ this $\delta$ works, and $\lim_{x \to 3} (2x - 1) = 5$ .

\boxed{\delta = \frac{\eps}{2}}

(6)

Check. For $\eps = 0.1$ : $\delta = 0.05$ , and $x = 3.04$ gives $\abs{2(3.04) - 1 - 5} = 0.08 < 0.1$ . ✓ The point $x = 3.05$ , just outside the window, gives exactly 0.1: with this function $\delta = \frac{\eps}{2}$ is the largest $\delta$ that works.

Example 2 (The stone’s speed at

t = 1

)

Show that the average speeds $v(h) = \dfrac{5(1+h)^2 - 5}{h}$ of Why this matters satisfy $\lim_{h \to 0} v(h) = 10$ , and find how small $\abs{h}$ must be for $v(h)$ to be within 0.01 m/s of 10 m/s.

Simplify. $v$ is defined for every $h \ne 0$ , which is all the definition looks at. For $h \ne 0$ ,
$v(h) = \frac{5(1 + 2h + h^2) - 5}{h} = \frac{10h + 5h^2}{h} = 10 + 5h .$
(7)
Scratch work. $\abs{v(h) - 10} = \abs{5h} = 5\abs{h}$ .
Choose $\delta$ and prove. Let $\eps > 0$ and $\delta = \frac{\eps}{5}$ . If $0 < \abs{h} < \delta$ , then $\abs{v(h) - 10} = 5\abs{h} < 5 \cdot \frac{\eps}{5} = \eps$ . So $\lim_{h \to 0} v(h) = 10$ .
The tolerance 0.01 m/s. By step 2, $\abs{v(h) - 10} < 0.01$ exactly when $5\abs{h} < 0.01$ , that is, when $\abs{h} < 0.002$ . So every time interval shorter than 0.002 s (on either side of $t = 1$ ) gives an average speed within 0.01 m/s of 10 m/s, and no longer one does: at $\abs{h} = 0.002$ the difference is exactly 0.01.

\boxed{\lim_{h \to 0} v(h) = 10 \text{ m/s}, \quad \abs{h} < 0.002 \text{ s}}

(8)

Check. $h = 0.001$ : $v(0.001) = \frac{5 \cdot 1.002001 - 5}{0.001} = 10.005$ , within 0.01 of 10. ✓ The table in the figure shows $v(0.01) = 10.05 = 10 + 5(0.01)$ . ✓ Units: metres divided by seconds. ✓

When a limit does not exist¶

To show that a number $L$ is not the limit, one round of the game that cannot be won is enough: one $\eps > 0$ such that every window $0 < \abs{x - a} < \delta$ , however small, contains a point $x$ with $\abs{f(x) - L} \ge \eps$ (or where $f$ is undefined). To show that the limit does not exist, we must do this for every real number $L$ . The next three examples show the three typical ways this happens: a jump, oscillation, and unbounded values. Each proof uses the same tool: two points of one window whose values are too far apart to be close to the same number $L$ .

Example 3 (A jump:

\lim_{x \to 0} \frac{\abs{x}}{x}

does not exist)

Let $f(x) = \dfrac{\abs{x}}{x}$ for $x \ne 0$ . Then $f(x) = 1$ for $x > 0$ and $f(x) = -1$ for $x < 0$ : the graph jumps from height -1 to height 1 at 0.

Strategy. Every window around 0 contains points on both sides of 0, where the values are -1 and 1, a distance 2 apart. No number is within 1 of both.
Proof. Let $L$ be any real number, and suppose that $\delta > 0$ works for $\eps = 1$ . The points $x = \frac{\delta}{2}$ and $x = -\frac{\delta}{2}$ are in the window, so $\abs{1 - L} < 1$ and $\abs{-1 - L} < 1$ , and $\abs{L - (-1)} = \abs{-1 - L}$ by property 2 of the absolute value. By part (b) of the triangle inequality,
$2 = \abs{1 - (-1)} \le \abs{1 - L} + \abs{L - (-1)} < 1 + 1 = 2,$
(9)
which is impossible. So no $\delta$ works for $\eps = 1$ , and $L$ is not the limit.
Since $L$ was arbitrary, no real number is the limit.

\boxed{\lim_{x \to 0} \frac{\abs{x}}{x} \text{ does not exist}}

(10)

Check. $f(0.001) = 1$ and $f(-0.001) = -1$ : arbitrarily close to 0 the values are still 2 apart. ✓ The proof never used $f(0)$ , which is undefined. ✓

Example 4 (Oscillation:

\lim_{x \to 0} \sin(1/x)

does not exist)

Let $f(x) = \sin(1/x)$ for $x \ne 0$ . As $x$ decreases to 0, $1/x$ increases without bound, so it passes again and again through multiples of $\pi$ , where $\sin$ is 0, and through numbers $\frac{\pi}{2} + 2k\pi$ , where $\sin$ is 1, by the facts from school.

Strategy. In every window around 0 we find a point where $\sin(1/x) = 0$ and a point where $\sin(1/x) = 1$ . No number is within $\frac12$ of both.
The points. For an integer $n \ge 1$ , let
$p_n = \frac{1}{2\pi n} \qquad\text{and}\qquad q_n = \frac{1}{2\pi n + \frac{\pi}{2}} .$
(11)
Then $f(p_n) = \sin(2\pi n) = 0$ and $f(q_n) = \sin\bigl(2\pi n + \frac{\pi}{2}\bigr) = 1$ , and $0 < q_n < p_n$ .
Proof. Let $L$ be any real number, and suppose that $\delta > 0$ works for $\eps = \frac12$ . By the Archimedean property, there is an integer $n \ge 1$ with $n > \frac{1}{2\pi\delta}$ . Then $2\pi n > \frac{1}{\delta}$ , so $0 < q_n < p_n = \frac{1}{2\pi n} < \delta$ : both points are in the window. Hence $\abs{0 - L} < \frac12$ and $\abs{1 - L} < \frac12$ , and $\abs{L - 0} = \abs{0 - L}$ by property 2 of the absolute value. By part (b) of the triangle inequality,
$1 = \abs{1 - 0} \le \abs{1 - L} + \abs{L - 0} < \tfrac12 + \tfrac12 = 1,$
(12)
which is impossible. So $L$ is not the limit, for any real $L$ .

\boxed{\lim_{x \to 0} \sin(1/x) \text{ does not exist}}

(13)

Check. For $n = 50$ : $p_{50} = \frac{1}{100\pi} \approx 0.003183$ gives $\sin(100\pi) = 0$ , and $q_{50} = \frac{1}{100\pi + \pi/2} \approx 0.003167$ gives $\sin\bigl(100\pi + \frac{\pi}{2}\bigr) = 1$ . Two points less than 0.0001 apart, values 1 apart. ✓

Example 5 (Unbounded values:

\lim_{x \to 0} \frac{1}{x^2}

does not exist)

Let $f(x) = \dfrac{1}{x^2}$ for $x \ne 0$ . Close to 0 the values are huge: $f(0.01) = 10\,000$ and $f(0.001) = 1\,000\,000$ .

Strategy. Whatever $L$ is proposed, every window around 0 contains a point where $f(x)$ is more than 1 above $L$ .
Proof. Let $L$ be any real number, and suppose that $\delta > 0$ works for $\eps = 1$ . Let $x = \min\bigl(\frac{\delta}{2}, \frac{1}{\abs{L} + 2}\bigr)$ . Then $0 < x < \delta$ , so $x$ is in the window. Also $x \le \frac{1}{\abs{L} + 2} \le \frac12 < 1$ ; multiplying $x < 1$ by $x > 0$ gives $x^2 < x$ , and so
$f(x) = \frac{1}{x^2} > \frac{1}{x} \ge \abs{L} + 2 .$
(14)
Then $f(x) - L \ge f(x) - \abs{L} > 2$ , so $\abs{f(x) - L} > 2 > 1 = \eps$ , a contradiction. So $L$ is not the limit, for any real $L$ .

\boxed{\lim_{x \to 0} \frac{1}{x^2} \text{ does not exist}}

(15)

Check. For $L = 100$ the proof takes $x \le \frac{1}{102}$ , where $f(x) \ge 102^2 = 10\,404$ , far more than 1 away from 100. ✓

Common mistakes¶

“The limit is just

f(a)

”

✗ Wrong: “ $h(1) = 5$ , so $\lim_{x \to 1} h(x) = 5$ ”, for the function $h$ of Remark 1, with $h(x) = x + 1$ for $x \ne 1$ and $h(1) = 5$ .

Why: the definition only looks at $x \ne a$ . The value $f(a)$ may be different from the limit, or not be defined at all, as for $\frac{x^2 - 1}{x - 1}$ at 1.

✓ Right: $\lim_{x \to 1} h(x) = 2$ , because $\abs{h(x) - 2} = \abs{x - 1}$ for $x \ne 1$ , so $\delta = \eps$ works. Substituting $x = a$ gives the limit only for functions that are known to behave well at $a$ , and that is something to prove, not assume.

Checking one

\eps

, or a few points

✗ Wrong: “For $\eps = 0.1$ the widget found $\delta = 0.0248$ , so $\lim_{x \to 2} x^2 = 4$ .” Or: “ $\sin(\pi/x)$ is 0 at $x = 0.1, 0.01, 0.001$ , so its limit at 0 is 0.”

Why: the definition says for every $\eps > 0$ , and every $x$ in the window. One $\eps$ , or finitely many points, is evidence, not a proof. The table of $\sin(\pi/x)$ in Why this matters is correct and still misleading: the limit does not exist (Exercise 9).

✓ Right: prove the inequality for an arbitrary $\eps > 0$ and an arbitrary $x$ in the window, as in Example 1. Use tables and the widget to guess the limit and the $\delta$ , then prove the guess.

Rigorous track¶

Example 6 (Prove

\lim_{x \to 2} x^2 = 4

from the definition)

Goal. Given $\eps > 0$ , find $\delta > 0$ such that $0 < \abs{x - 2} < \delta$ implies $\abs{x^2 - 4} < \eps$ . The function $x^2$ is defined for every $x$ .

Scratch work. $\abs{x^2 - 4} = \abs{x - 2}\,\abs{x + 2}$ . The first factor is the one that $\delta$ controls. The second, $\abs{x + 2}$ , changes with $x$ , so we first bound it on a fixed window.
Bound the other factor. If $\abs{x - 2} < 1$ , then $1 < x < 3$ , by part (a) of the proposition on distance inequalities, so $3 < x + 2 < 5$ and $\abs{x + 2} < 5$ .
Choose $\delta$ . On that window, $\abs{x^2 - 4} < 5\abs{x - 2}$ , which is less than $\eps$ when $\abs{x - 2} < \frac{\eps}{5}$ . Both conditions hold if $\delta = \min\bigl(1, \frac{\eps}{5}\bigr)$ .
Proof. Let $\eps > 0$ and $\delta = \min\bigl(1, \frac{\eps}{5}\bigr)$ , and let $0 < \abs{x - 2} < \delta$ . Since $\delta \le 1$ , step 2 gives $\abs{x + 2} < 5$ . Since $\delta \le \frac{\eps}{5}$ and $\abs{x - 2} > 0$ ,
$\abs{x^2 - 4} = \abs{x - 2}\,\abs{x + 2} < 5\abs{x - 2} < 5 \cdot \frac{\eps}{5} = \eps .$
(16)
(The first $<$ multiplies $\abs{x + 2} < 5$ by the positive number $\abs{x - 2}$ .) So $\lim_{x \to 2} x^2 = 4$ .

\boxed{\delta = \min\Bigl(1, \frac{\eps}{5}\Bigr)}

(17)

Why the 1? It caps the window, and that is what makes $\abs{x + 2} < 5$ true. Without it, $\delta = \frac{\eps}{5}$ fails for large $\eps$ : for $\eps = 100$ it gives $\delta = 20$ , and $x = 21$ is in the window but $\abs{21^2 - 4} = 437 > 100$ . Any fixed cap would do; a cap of $\frac12$ gives $\abs{x + 2} < 4.5$ and $\delta = \min\bigl(\frac12, \frac{\eps}{4.5}\bigr)$ .

Check. For $\eps = 0.5$ : $\delta = \min(1, 0.1) = 0.1$ , below the largest $\delta$ that the ε–δ figure reports for both sides, 0.12132. ✓ The point $x = 2.09$ gives $\abs{2.09^2 - 4} = 0.3681 < 0.5$ . ✓

The definition in symbols, and its negation

With quantifiers, $\lim_{x \to a} f(x) = L$ says

\forall \eps > 0 \ \ \exists \delta > 0 \ \ \forall x \colon \quad 0 < \abs{x - a} < \delta \implies \bigl(x \in \dom f \text{ and } \abs{f(x) - L} < \eps\bigr).

(18)

Here $\dom f$ is the domain of $f$ . The order matters. “ $\exists \delta > 0 \ \forall \eps > 0$ ” would ask for one window that works for every tolerance at once. For $f(x) = x$ at $a = 0$ , with $L = 0$ , that is false: for each $\delta > 0$ , the tolerance $\eps = \frac{\delta}{2}$ fails at the point $x = \frac{3\delta}{4}$ of the window, where $\abs{f(x) - 0} = \frac{3\delta}{4} \ge \frac{\delta}{2}$ .

To negate, each $\forall$ becomes $\exists$ and each $\exists$ becomes $\forall$ , the implication $P \implies Q$ becomes “ $P$ and not $Q$ ”, and, by De Morgan’s law, “not ( $A$ and $B$ )” becomes “(not $A$ ) or (not $B$ )”. So $L$ is not the limit when

\exists \eps > 0 \ \ \forall \delta > 0 \ \ \exists x \colon \quad 0 < \abs{x - a} < \delta \ \text{ and } \ \bigl(x \notin \dom f \text{ or } \abs{f(x) - L} \ge \eps\bigr).

(19)

This is the pattern of the three non-existence examples: one $\eps$ , and for each $\delta$ a point $x$ that fails.

Why “

0 < \abs{x - a}

”?

The condition $0 < \abs{x - a}$ removes $x = a$ from every window. Without it, the definition would demand $\abs{f(a) - L} < \eps$ for every $\eps > 0$ , that is, $f(a) = L$ , and it would not apply at all where $f(a)$ is undefined. Both are exactly the cases we need limits for: the average speed $\frac{5(1+h)^2 - 5}{h}$ is undefined at $h = 0$ . The definition is built to describe how $f$ behaves near $a$ , which is information that $f(a)$ cannot give.

Why

f

must be defined near

a

Definition 1 asks for $f$ to be defined on an open interval around $a$ (except possibly at $a$ ), and, in its domain clause, for $f$ to be defined at every point of the window. The proof of Theorem 1 rests on that clause of the definition: it picks a point $x = a + \frac{\delta}{2}$ in the window and uses $f(x)$ .

Suppose the definition dropped both requirements, the open interval and the domain clause, and only asked that “ $x \in \dom f$ and $0 < \abs{x - a} < \delta$ imply $\abs{f(x) - L} < \eps$ ”. Take $f(x) = \sqrt{x}$ and $a = -1$ . For $\delta \le 1$ no $x$ in the domain $[0, \infty)$ satisfies $0 < \abs{x + 1} < \delta$ , so the condition holds for every $L$ , and every real number would be “the limit” of $\sqrt{x}$ at -1. Uniqueness fails, so that version of the definition needs a further hypothesis.

Many analysis books use such a domain-relative definition, with the hypothesis that every window around $a$ contains points of the domain other than $a$ ( $a$ is a limit point of the domain). It agrees with ours whenever $f$ is defined on an open interval around $a$ (except possibly at $a$ ), and it also allows endpoints: it gives $\lim_{x \to 0} \sqrt{x} = 0$ , where our definition does not apply, because $\sqrt{x}$ is undefined to the left of 0.

Summary¶

$\lim_{x \to a} f(x) = L$ is about the values of $f$ near $a$ , never at $a$ : $f(a)$ may differ from the limit or be undefined (Remark 1).
Precisely (Definition 1): for every $\eps > 0$ there is a $\delta > 0$ such that
$\boxed{0 < \abs{x - a} < \delta \implies \abs{f(x) - L} < \eps .}$
(20)
The $\delta$ may depend on $\eps$ and $a$ , never on $x$ . To prove a limit, rewrite $\abs{f(x) - L}$ in terms of $\abs{x - a}$ , choose $\delta$ , then check it; for a quadratic, cap the window first, as in $\delta = \min\bigl(1, \frac{\eps}{5}\bigr)$ .
A limit, if it exists, is unique (Theorem 1).
A limit fails to exist when, for every $L$ , some $\eps$ defeats every $\delta$ : typically a jump, oscillation, or unbounded values.
Tables, graphs and the ε–δ widget suggest limits and $\delta$ s; only a proof establishes them.

Exercises¶

Solution to Exercise 1 #

To six decimal places:

$x$	0.1	0.01	0.001	-0.001	-0.01	-0.1
$\frac{\sqrt{1 + x} - 1}{x}$	0.488088	0.498756	0.499875	0.500125	0.501256	0.513167

As $x$ approaches 0, the values seem to approach $\frac12$ : those from the right increase towards it and those from the left decrease towards it. At $x = \pm 0.001$ they agree to two decimal places, 0.50. So we estimate the limit as 0.50. This is an estimate, not a proof: a table checks only finitely many points (Why this matters).

Exercise 5 (The exact δ in the ε–δ figure)

In the ε–δ figure, $f(x) = x^2$ , $a = 2$ and $L = 4$ . For $\eps = 0.1$ , find exactly the largest $\delta > 0$ such that $\abs{x^2 - 4} < 0.1$ for every $x$ in the window (a) on the right, $2 < x < 2 + \delta$ ; (b) on the left, $2 - \delta < x < 2$ ; (c) on both sides, $0 < \abs{x - 2} < \delta$ . Then compare with the numbers the widget reports.

Hint 1

$\abs{x^2 - 4} < 0.1$ means $3.9 < x^2 < 4.1$ . For $x > 0$ , take square roots.

Hint 2

For (c), the window must fit inside the allowed region on both sides at once.

Answer

(a) $\frac{\sqrt{410}}{10} - 2$ (b) $2 - \frac{\sqrt{390}}{10}$ (c) $\frac{\sqrt{410}}{10} - 2$

Solution to Exercise 5 #

By part (a) of the proposition on distance inequalities, applied to $x^2$ with $a = 4$ and $\delta = 0.1$ , $\abs{x^2 - 4} < 0.1$ means $3.9 < x^2 < 4.1$ . For $x > 0$ , this means $\sqrt{3.9} < x < \sqrt{4.1}$ , by the order part of the square-root remark: for $s, t \ge 0$ , $s < t$ exactly when $s^2 < t^2$ , and here $(\sqrt{3.9})^2 = 3.9$ and $(\sqrt{4.1})^2 = 4.1$ . Every window below lies in $x > 0$ .

(a) For $x > 2$ the condition $x > \sqrt{3.9}$ holds automatically ( $\sqrt{3.9} < 2$ , by the same order part, since $3.9 < 4 = 2^2$ ), so we need $x < \sqrt{4.1}$ for every $x$ in $(2, 2 + \delta)$ . That holds exactly when $2 + \delta \le \sqrt{4.1}$ . The largest $\delta$ is $\sqrt{4.1} - 2 \approx 0.0248457$ .

(b) For $x < 2$ the condition $x < \sqrt{4.1}$ holds automatically ( $2 < \sqrt{4.1}$ , since $2^2 = 4 < 4.1$ ), so we need $x > \sqrt{3.9}$ for every $x$ in $(2 - \delta, 2)$ , that is, $2 - \delta \ge \sqrt{3.9}$ . The largest $\delta$ is $2 - \sqrt{3.9} \approx 0.0251582$ .

(c) The window works on both sides exactly when $\delta$ is at most both numbers, so the largest is the smaller one. To compare them without a calculator: $\sqrt{4.1} - 2 < 2 - \sqrt{3.9}$ means $\sqrt{4.1} + \sqrt{3.9} < 4$ . We compare squares, using the order part of the square-root remark for non-negative numbers. First, $\bigl(\sqrt{4.1}\,\sqrt{3.9}\bigr)^2 = 4.1 \cdot 3.9 = 15.99 < 16 = 4^2$ , so $\sqrt{4.1}\,\sqrt{3.9} < 4$ . Then

\bigl(\sqrt{4.1} + \sqrt{3.9}\bigr)^2 = 4.1 + 2\sqrt{4.1}\,\sqrt{3.9} + 3.9 < 8 + 2 \cdot 4 = 16 = 4^2,

(22)

so $\sqrt{4.1} + \sqrt{3.9} < 4$ . So the right side’s value $\sqrt{4.1} - 2$ is the smaller one, and it is the answer to (c).

The widget reports 0.0248456 and 0.0251582: here, these are the exact values rounded down to six significant digits. Its method promises less: the $\delta$ it reports passed at every point it checked and lies below every point where it saw the check fail, and it may be smaller than the exact value by about 10^-5 of that value.

In the answer, the square roots are written with whole numbers under the root: $\sqrt{4.1} = \frac{\sqrt{410}}{10}$ , because $\frac{\sqrt{410}}{10} \ge 0$ and $\bigl(\frac{\sqrt{410}}{10}\bigr)^2 = \frac{410}{100} = 4.1$ , and the square root is unique (the square-root remark). In the same way, $\sqrt{3.9} = \frac{\sqrt{390}}{10}$ .

Where this leads¶

These topics build on this one:

One-Sided Limits
Limit Laws
Tangent Lines and Rates of Change (coming soon)

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Every page of the Limits chapter plays the ε–δ game of this page: the limit laws are proved with it, once, so that later limits can be computed without it.