One-Sided Limits - Maths: an open, verified university mathematics reference

Why this matters¶

A courier charges £3 for a parcel of up to 2 kg, and £5 for a parcel over 2 kg and up to 10 kg. For a parcel of $w$ kilograms, where $0 < w \le 10$ , the price in pounds is

P(w) = \begin{cases} 3 & \text{if } 0 < w \le 2,\\ 5 & \text{if } 2 < w \le 10. \end{cases}

(1)

What does the price do near 2 kg? A parcel of 1.999 kg costs £3 and one of 2.001 kg costs £5, and however close to 2 kg the two weights are, the two prices stay £2 apart. So $P(w)$ approaches no single number as $w$ approaches 2: $\lim_{w \to 2} P(w)$ does not exist, as we prove in Example 2. But “does not exist” throws away what we know. Just below 2 kg the price is £3, and just above it is £5. We would like to say that $P(w)$ approaches 3 as $w$ approaches 2 from the left, approaches 5 from the right, and so jumps by £2 at 2 kg.

A second situation has the same cure. The definition of the limit asks for $f$ to be defined on an open interval around $a$ , except possibly at $a$ . The square root $\sqrt{x}$ is defined for no $x < 0$ , because a negative number has no real square root (the square-root remark), so that definition says nothing about $\sqrt{x}$ at 0. Yet $\sqrt{x}$ is as small as we like when $x > 0$ is small enough: $\sqrt{0.0001} = 0.01$ .

A one-sided limit looks at one side of $a$ only. It describes the jump of the price, and the square root at 0, and it gives a practical test for whether a limit exists: look at the two sides separately, then compare.

Limits from one side¶

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 $a$ and greater than $a$ . The window of the ε–δ game becomes the right half of the window of the definition of the limit: the punctured neighbourhood $0 < \abs{x - a} < \delta$ is the set $(a - \delta, a) \cup (a, a + \delta)$ , by part (e) of the proposition on distance inequalities, and a one-sided limit keeps only one of the two intervals.

Definition 1 (One-sided limits)

Let $a$ and $L$ be real numbers.

(a) Let $f$ be a function that is defined at every point of an open interval $(a, a + r)$ , for some $r > 0$ . We say that the limit of $f(x)$ as $x$ approaches $a$ from the right is $L$ , and write

\lim_{x \to a^{+}} f(x) = L,

(2)

if for every $\eps > 0$ there is a $\delta > 0$ such that $f$ is defined at every $x$ with $a < x < a + \delta$ , and

\begin{aligned} &a < x < a + \delta \\ &\quad\text{implies}\quad \abs{f(x) - L} < \eps . \end{aligned}

(3)

(b) Let $f$ be a function that is defined at every point of an open interval $(a - r, a)$ , for some $r > 0$ . We say that the limit of $f(x)$ as $x$ approaches $a$ from the left is $L$ , and write

\lim_{x \to a^{-}} f(x) = L,

(4)

if for every $\eps > 0$ there is a $\delta > 0$ such that $f$ is defined at every $x$ with $a - \delta < x < a$ , and

\begin{aligned} &a - \delta < x < a \\ &\quad\text{implies}\quad \abs{f(x) - L} < \eps . \end{aligned}

(5)

These are the right-hand and the left-hand limit, together the one-sided limits. We also write $f(x) \to L$ as $x \to a^{+}$ (or $x \to a^{-}$ ).

In words. The right-hand limit is the definition of the limit with its window $0 < \abs{x - a} < \delta$ replaced by the right half, $a < x < a + \delta$ ; the left-hand limit keeps the left half, $a - \delta < x < a$ . The game is the same: a challenger names $\eps > 0$ , you reply with $\delta > 0$ , and you win the round if every $x$ in your half-window has $f(x)$ within $\eps$ of $L$ . Three points to notice:

Neither half-window contains $a$ . The value $f(a)$ , and whether it exists, plays no part, for the same reason as in the remark that the value at $a$ does not matter.
The values of $f$ on the other side of $a$ play no part either. For the right-hand limit, $f$ need not even be defined to the left of $a$ : that is what makes $\lim_{x \to 0^{+}} \sqrt{x}$ meaningful.
As before, if a $\delta$ wins a round, so does every smaller $\delta > 0$ . Since $f$ is defined on $(a, a + r)$ (in part (a)), every $\delta \le r$ gives a half-window inside that interval, so the requirement that $f$ be defined in the window only rules out windows that are too wide.

Example. Let $g(x) = x$ for $x \le 0$ and $g(x) = x + 1$ for $x > 0$ . Then $\lim_{x \to 0^{+}} g(x) = 1$ . Given $\eps > 0$ , take $\delta = \eps$ . If $0 < x < \delta$ , then $g(x) = x + 1$ , so $\abs{g(x) - 1} = \abs{x} = x < \eps$ , where $\abs{x} = x$ because $x > 0$ (the definition of the absolute value).

Non-example. $\lim_{x \to 0^{+}} g(x)$ is not 0, although $g(0) = 0$ . Take $\eps = \frac12$ . Whatever $\delta > 0$ is offered, let $d = \frac12 \min\bigl(\delta, 1\bigr)$ , which is positive. Since $\min(\delta, 1) \le \delta$ , multiplying by the positive number $\frac12$ keeps the inequality, so $d \le \frac{\delta}{2} < \delta$ , and the half-window $0 < x < \delta$ contains $x = d$ . There $g(d) = d + 1$ , and adding 1 to $d > 0$ keeps the inequality, $d + 1 > 1$ ; so $d + 1$ is positive, and $\abs{g(d) - 0} = d + 1 > 1 > \frac12$ by the definition of the absolute value. So no $\delta$ wins the round $\eps = \frac12$ . (From the left, $g(x) = x$ , and $\lim_{x \to 0^{-}} g(x) = 0$ , with $\delta = \eps$ as in the example.)

The figure plays the game of the definition of the limit for the parcel price of Why this matters, with the two-sided claim $\lim_{w \to 2} P(w) = 5$ . It reports on the two sides of 2 separately, the left half-window and the right half-window, which is exactly what the one-sided limits do.

Figure 1:The ε–δ game for the claim that the parcel price has the limit 5 at 2 kg. Graph of the price for weights from 1 kg to 3 kg: height 3 to the left of 2 and height 5 to the right, with an open circle at $(2, 5)$ , the claimed limit. Around it are a horizontal band of half-width $\eps$ around $y = 5$ (translucent, dashed edges) and a vertical window $0 < \abs{x - 2} < \delta$ (translucent, solid edges; the dotted line $x = 2$ itself is left out, so the widget never uses the price at exactly 2 kg). A slider sets $\eps$ from 0.1 to 2.5 in steps of 0.1, starting at 0.5. For each $\eps$ the widget states, separately for the left and the right of 2, the largest $\delta$ it found or that it observed none. For $\eps = 0.5$ it reports on the right that the values stayed in the band at every point it checked, out to the edge of the view ( $\delta = 1$ ), and on the left that it observed no $\delta$ : the values leave the band at points as close to 2 as it checks. A second slider sets your own $\delta$ ; 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:

At $\eps = 0.5$ , read what the widget reports on each side of 2. Which side’s values are within $\eps$ of 5?
Select the $\eps$ slider and raise $\eps$ one step at a time. At which $\eps$ does the widget first find a $\delta$ on the left? Why there, and not one step earlier?
Now the widget finds a $\delta$ on both sides. Does that make 5 the limit of $P(w)$ as $w \to 2$ ? (The definition asks for a $\delta$ for every $\eps > 0$ .)

The widget checks finitely many points, so what it reports are observations. The proofs are on this page: Example 2 proves that the right-hand limit is 5 and the left-hand limit is 3, so that $\lim_{w \to 2} P(w)$ does not exist, and exercise Exercise 8 finds exactly the $\eps$ for which some $\delta$ works on the left.

Could a function have two different right-hand limits at the same point? No, by the argument of the uniqueness of limits, now with a point taken in the half-window.

Remark 1 (One-sided limits are unique)

Let $f$ and $a$ be as in part (a) of Definition 1. If $\lim_{x \to a^{+}} f(x) = L$ and $\lim_{x \to a^{+}} f(x) = M$ , then $L = M$ . The same holds for left-hand limits, with $f$ and $a$ as in part (b).

Reason. Suppose $L \ne M$ , and let $\eps = \frac12 \abs{L - M}$ , which is positive by property 1 of the absolute value, since $L - M \ne 0$ . By part (a) of the definition, applied to $L$ and to $M$ , there are $\delta_1 > 0$ and $\delta_2 > 0$ such that $f$ is defined at every $x$ with $a < x < a + \delta_1$ and has $\abs{f(x) - L} < \eps$ there, and is defined at every $x$ with $a < x < a + \delta_2$ and has $\abs{f(x) - M} < \eps$ there. Let $\delta = \min(\delta_1, \delta_2)$ , which is positive, and $x = a + \frac{\delta}{2}$ . Since $0 < \frac{\delta}{2} < \delta$ , adding $a$ keeps both inequalities: $a < x < a + \delta$ . Since $\delta \le \delta_1$ , adding $a$ gives $a + \delta \le a + \delta_1$ , so $a < x < a + \delta_1$ by transitivity; in the same way $a < x < a + \delta_2$ . So $x$ lies in both half-windows: $f(x)$ is defined, $\abs{f(x) - L} < \eps$ and $\abs{f(x) - M} < \eps$ . By property 2 of the absolute value, $\abs{L - f(x)} = \abs{f(x) - L}$ . By part (b) of the triangle inequality, $\abs{u - v} \le \abs{u - w} + \abs{w - v}$ , with $u = L$ , $v = M$ and $w = f(x)$ ,

\begin{aligned} \abs{L - M} &\le \abs{L - f(x)} \\ &\quad + \abs{f(x) - M} \\ &< \eps + \eps = \abs{L - M}, \end{aligned}

(6)

which is impossible. So $L = M$ . For left-hand limits, take $x = a - \frac{\delta}{2}$ instead. Multiplying $0 < \frac{\delta}{2} < \delta$ by the negative number ${-1}$ reverses both inequalities, $-\delta < -\frac{\delta}{2} < 0$ , and adding $a$ keeps them: $a - \delta < x < a$ . Multiplying $\delta \le \delta_1$ by ${-1}$ reverses it and adding $a$ keeps it, $a - \delta_1 \le a - \delta$ , so $x$ lies in the half-window of $\delta_1$ , and in the same way in that of $\delta_2$ .

So we may speak of the right-hand limit and the left-hand limit, when they exist.

Main results¶

The two half-windows together make up the punctured neighbourhood $0 < \abs{x - a} < \delta$ . So a limit should exist exactly when $f(x)$ settles near the same number on both sides.

Proof (Rigorous track)

By part (e) of the proposition on distance inequalities, the window $0 < \abs{x - a} < \delta$ is made of the two half-windows $a - \delta < x < a$ and $a < x < a + \delta$ . So a $\delta$ that works for the limit works on each side, and conversely the smaller of a $\delta$ for each side works for the limit.

The one-sided limits make sense. By hypothesis, $f$ is defined at every point of an open interval $I$ containing $a$ , except possibly at $a$ . By the definition of an interval, $I$ is $(c, d)$ , $(c, \infty)$ , $(-\infty, d)$ or $\R$ , and since $a \in I$ , we have $c < a$ whenever $I$ has a left endpoint $c$ , and $a < d$ whenever it has a right endpoint $d$ . Let $r$ be the smaller of the numbers $a - c$ and $d - a$ that occur (one of them, if $I$ has only one endpoint), and $r = 1$ if $I = \R$ . Then $r > 0$ : adding $-c$ to $c < a$ keeps the inequality, $0 < a - c$ , and adding $-a$ to $a < d$ gives $0 < d - a$ . Every $x$ with $a - r < x < a + r$ lies in $I$ . Adding $c - r$ to $r \le a - c$ keeps the inequality, $c \le a - r$ , so $c < x$ by transitivity; and adding $a$ to $r \le d - a$ gives $a + r \le d$ , so $x < d$ by transitivity. So $f$ is defined at every point of $(a - r, a)$ and of $(a, a + r)$ , as parts (b) and (a) of Definition 1 require.

If the limit is $L$ , so are both one-sided limits. Let $\eps > 0$ . Since $\lim_{x \to a} f(x) = L$ , there is a $\delta > 0$ such that $f$ is defined at every $x$ with $0 < \abs{x - a} < \delta$ , and $\abs{f(x) - L} < \eps$ for each such $x$ . If $a < x < a + \delta$ , then $0 < \abs{x - a} < \delta$ , by part (e) of the proposition, read from right to left. So $f(x)$ is defined and $\abs{f(x) - L} < \eps$ : this $\delta$ wins the round $\eps$ for the right-hand limit. In the same way, $a - \delta < x < a$ implies $0 < \abs{x - a} < \delta$ , so the same $\delta$ wins the round for the left-hand limit. Since $\eps > 0$ was arbitrary, $\lim_{x \to a^{+}} f(x) = L$ and $\lim_{x \to a^{-}} f(x) = L$ .

If both one-sided limits are $L$ , so is the limit. Let $\eps > 0$ . By part (b) of the definition there is a $\delta_1 > 0$ such that $f$ is defined and $\abs{f(x) - L} < \eps$ at every $x$ with $a - \delta_1 < x < a$ ; by part (a) there is a $\delta_2 > 0$ such that the same holds at every $x$ with $a < x < a + \delta_2$ . Let $\delta = \min(\delta_1, \delta_2)$ , which is positive, and let $0 < \abs{x - a} < \delta$ . By part (e) of the proposition, read from left to right, either $a - \delta < x < a$ or $a < x < a + \delta$ .

In the first case, multiplying $\delta \le \delta_1$ by the negative number ${-1}$ reverses it, $-\delta_1 \le -\delta$ , and adding $a$ keeps it: $a - \delta_1 \le a - \delta < x < a$ . So $x$ is in the half-window of $\delta_1$ .
In the second case, adding $a$ to $\delta \le \delta_2$ gives $a < x < a + \delta \le a + \delta_2$ , so $x$ is in the half-window of $\delta_2$ .

Either way, $f(x)$ is defined and $\abs{f(x) - L} < \eps$ . So $\delta$ wins the round $\eps$ for the limit; since $\eps > 0$ was arbitrary, $\lim_{x \to a} f(x) = L$ .

The theorem gives the limit when the two one-sided limits agree. Read the other way, it is the usual way to show that a limit does not exist.

Proof

Both parts argue by contradiction from the “only if” part of Theorem 1. Suppose that $\lim_{x \to a} f(x) = N$ for some real number $N$ . By that part of the theorem, $\lim_{x \to a^{-}} f(x) = N$ and $\lim_{x \to a^{+}} f(x) = N$ .

(a) From the left, $f$ has the left-hand limits $N$ and $L$ , so $N = L$ by Remark 1. From the right, it has the right-hand limits $N$ and $M$ , so $N = M$ by the same remark. Then $L = N = M$ , which contradicts $L \ne M$ .

(b) Both one-sided limits exist, since both are $N$ ; this contradicts the hypothesis that one of them does not exist.

In both cases no such $N$ exists, so $\lim_{x \to a} f(x)$ does not exist.

How to use them. To decide whether $\lim_{x \to a} f(x)$ exists, find both one-sided limits and compare them. If they are equal, the limit is their common value, by Theorem 1. If they are different, or one of them does not exist, there is no limit, by part (a) or part (b) of Corollary 1.

Worked examples¶

Piecewise functions¶

A function given by different formulas on the two sides of $a$ is made for one-sided limits: each half-window sees only one formula. The value at $a$ itself, whichever formula or special value gives it, plays no part.

Example 1 (Two formulas that meet, and a value that does not)

Let

f(x) = \begin{cases} x^2 & \text{if } x < 2,\\ 0 & \text{if } x = 2,\\ 2x & \text{if } x > 2. \end{cases}

(9)

Find $\lim_{x \to 2^{-}} f(x)$ and $\lim_{x \to 2^{+}} f(x)$ , and decide whether $\lim_{x \to 2} f(x)$ exists.

The value at 2. $f(2) = 0$ , given by the middle line. Neither half-window contains 2, so this value plays no part in either one-sided limit. The function is defined at every real number.
From the left: scratch work. For $x < 2$ , $f(x) = x^2$ , and $\abs{x^2 - 4} = \abs{x - 2}\,\abs{x + 2}$ by property 4 of the absolute value. The factor $\abs{x + 2}$ changes with $x$ , so we bound it on a fixed half-window: if $1 < x < 2$ , then adding 2 keeps the inequalities, $3 < x + 2 < 4$ , so $x + 2$ is positive and $\abs{x + 2} = x + 2 < 4$ (the definition of the absolute value). On that half-window $\abs{x^2 - 4} < 4\abs{x - 2}$ , which is less than $\eps$ when $\abs{x - 2} < \frac{\eps}{4}$ . Both conditions hold if $\delta = \min\bigl(1, \frac{\eps}{4}\bigr)$ .
From the left: proof. Let $\eps > 0$ , $\delta = \min\bigl(1, \frac{\eps}{4}\bigr)$ , and $2 - \delta < x < 2$ . Since $\delta \le 1$ , multiplying by ${-1}$ reverses the inequality and adding 2 keeps it, so $2 - \delta \ge 1$ and $1 < x < 2$ ; by step 2, $\abs{x + 2} < 4$ . Since $x < 2$ , $\abs{x - 2} = 2 - x$ , which is positive and less than $\delta \le \frac{\eps}{4}$ . Multiplying $\abs{x + 2} < 4$ by the positive number $\abs{x - 2}$ , and then $\abs{x - 2} < \frac{\eps}{4}$ by the positive number 4, keeps both strict inequalities:
$\begin{aligned} \abs{f(x) - 4} &= \abs{x - 2}\,\abs{x + 2} \\ &< 4\abs{x - 2} \\ &< 4 \cdot \frac{\eps}{4} = \eps . \end{aligned}$
(10)
So $\lim_{x \to 2^{-}} f(x) = 4$ .
From the right. For $x > 2$ , $f(x) = 2x$ , and $\abs{2x - 4} = 2\abs{x - 2} = 2(x - 2)$ , by property 4 of the absolute value and because $x - 2 > 0$ . Let $\eps > 0$ and $\delta = \frac{\eps}{2}$ . If $2 < x < 2 + \delta$ , then $0 < x - 2 < \frac{\eps}{2}$ , and multiplying by the positive number 2 keeps the inequality: $\abs{f(x) - 4} = 2(x - 2) < \eps$ . So $\lim_{x \to 2^{+}} f(x) = 4$ .
The limit. $f$ is defined on the open interval $\R$ around 2, and both one-sided limits are 4. By Theorem 1, $\lim_{x \to 2} f(x) = 4$ , although $f(2) = 0$ .

\boxed{ \begin{aligned} \lim_{x \to 2^{-}} f(x) &= \lim_{x \to 2^{+}} f(x) = 4, \\ \lim_{x \to 2} f(x) &= 4 \end{aligned} }

(11)

Check. For $\eps = 0.4$ , the left $\delta$ is $\min(1, 0.1) = 0.1$ , and $x = 1.95$ gives $\abs{1.95^2 - 4} = \abs{3.8025 - 4} = 0.1975 < 0.4$ . ✓ The right $\delta$ is 0.2, and $x = 2.15$ gives $\abs{4.3 - 4} = 0.3 < 0.4$ . ✓ Neither check used $f(2)$ . ✓ On the left the cap $\delta \le 1$ gives the bound $\abs{x + 2} < 4$ ; the two-sided window $0 < \abs{x - 2} < 1$ would only give $\abs{x + 2} < 5$ , as in the quadratic ε–δ example. ✓

Example 2 (The parcel price at 2 kg)

For the price $P(w)$ of Why this matters, find the one-sided limits at $w = 2$ , decide whether $\lim_{w \to 2} P(w)$ exists, and find the jump in price.

The value at 2. $P(2) = 3$ , from the first line of the price list ( $0 < w \le 2$ ). It plays no part in either one-sided limit, because neither half-window contains 2. $P$ is defined on the open interval $(0, 10)$ , which contains 2.
From the left. For $1 < w < 2$ , $P(w) = 3$ (first line). Let $\eps > 0$ and $\delta = 1$ . If $2 - \delta < w < 2$ , that is, $1 < w < 2$ , then $\abs{P(w) - 3} = 0 < \eps$ . So $\lim_{w \to 2^{-}} P(w) = 3$ . This 3 comes from the prices of parcels lighter than 2 kg; that $P(2)$ is also 3 is a coincidence of the price list, and the limit would be the same if the first line read $0 < w < 2$ .
From the right. For $2 < w < 3$ , $P(w) = 5$ (second line). Let $\eps > 0$ and $\delta = 1$ . If $2 < w < 2 + \delta$ , then $\abs{P(w) - 5} = 0 < \eps$ . So $\lim_{w \to 2^{+}} P(w) = 5$ .
The limit. The one-sided limits exist and are different, $3 \ne 5$ . By part (a) of Corollary 1, $\lim_{w \to 2} P(w)$ does not exist.
The jump. By Remark 2, the jump is $\lim_{w \to 2^{+}} P(w) - \lim_{w \to 2^{-}} P(w) = 5 - 3 = 2$ : the price jumps by £2 at 2 kg.

\boxed{ \begin{aligned} \lim_{w \to 2^{-}} P(w) &= 3, \\ \lim_{w \to 2^{+}} P(w) &= 5, \\ \text{jump} &= 2 \end{aligned} }

(12)

Check. $P(1.999) = 3$ and $P(2.001) = 5$ , in pounds: the prices on the two sides stay 2 apart however close to 2 kg the weights are. ✓ The widget in the figure found a $\delta$ on the right for $\eps = 0.5$ , and observed none on the left, where every value is 3, at distance 2 from 5. ✓ Units: pounds throughout. ✓

The jump of $\frac{\abs{x}}{x}$ at 0, the first of the three ways a limit can fail to exist in The Limit of a Function, fits the same pattern: its one-sided limits are ${-1}$ and 1 (exercise Exercise 1), so part (a) of Corollary 1 gives again what the jump example proved there with the triangle inequality.

An endpoint¶

Example 3 (

\lim_{x \to 0^{+}} \sqrt{x} = 0

)

The square root is defined on $[0, \infty)$ and on no negative number (the square-root remark). So the definition of the limit does not apply at 0, but part (a) of Definition 1 does: $\sqrt{x}$ is defined at every point of the open interval $(0, 1)$ , and every half-window $0 < x < \delta$ lies in its domain.

Goal. Given $\eps > 0$ , find $\delta > 0$ such that $0 < x < \delta$ implies $\abs{\sqrt{x} - 0} < \eps$ .

Scratch work. For $x > 0$ , $\sqrt{x} \ge 0$ , so $\abs{\sqrt{x} - 0} = \sqrt{x}$ (the definition of the absolute value). Both $\sqrt{x}$ and $\eps$ are non-negative, so by the order part of the square-root remark, $\sqrt{x} < \eps$ exactly when $(\sqrt{x})^2 < \eps^2$ , that is, when $x < \eps^2$ .
Choose $\delta$ . Take $\delta = \eps^2$ , which is positive.
Proof. Let $\eps > 0$ and $\delta = \eps^2$ . If $0 < x < \delta$ , then $\sqrt{x}$ is defined and $(\sqrt{x})^2 = x < \eps^2$ , so $\sqrt{x} < \eps$ by step 1, and $\abs{\sqrt{x} - 0} = \sqrt{x} < \eps$ . So $\lim_{x \to 0^{+}} \sqrt{x} = 0$ .
The other side. $\sqrt{x}$ is defined at no point of any interval $(-r, 0)$ , so part (b) of the definition does not apply: $\lim_{x \to 0^{-}} \sqrt{x}$ is not defined, and neither is $\lim_{x \to 0} \sqrt{x}$ . Theorem 1 does not apply either: its hypothesis asks for $f$ to be defined on both sides of $a$ .

\boxed{\delta = \eps^2}

(13)

Check. For $\eps = 0.1$ : $\delta = 0.01$ , and $x = 0.0081$ gives $\sqrt{x} = 0.09 < 0.1$ . ✓ The point $x = 0.01$ , just outside the half-window, gives exactly 0.1: no larger $\delta$ works, because every larger half-window contains $x = \eps^2$ , where $\sqrt{x} = \eps$ . ✓

Common mistakes¶

Rigorous track¶

One-sided limits in symbols, and the negation

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

\begin{aligned} &\forall \eps > 0 \ \ \exists \delta > 0 \ \ \forall x \colon \\ &\quad a < x < a + \delta \implies \\ &\qquad x \in \dom f \\ &\qquad \text{and } \abs{f(x) - L} < \eps . \end{aligned}

(14)

Here $\dom f$ is the domain of $f$ . To negate, each $\forall$ becomes $\exists$ and each $\exists$ becomes $\forall$ , the implication $P \implies Q$ becomes “ $P$ and not $Q$ ”, “not ( $A$ and $B$ )” becomes “(not $A$ ) or (not $B$ )”, and “not ( $u < v$ )” becomes “ $u \ge v$ ”, by trichotomy (the order rules of Absolute Value and Inequalities, “Working with inequalities”). So $L$ is not the right-hand limit when

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

(15)

This is the pattern of the non-example after Definition 1: $\eps = \frac12$ , and for each $\delta$ the point $x = \frac12 \min(\delta, 1)$ . For the left-hand limit, replace $a < x < a + \delta$ by $a - \delta < x < a$ .

Why

f

must be defined on one side of

a

Part (a) of Definition 1 asks for $f$ to be defined on an interval $(a, a + r)$ , and, in its domain clause, at every point of the half-window. The reason in Remark 1 rests on that clause: it picks the point $x = a + \frac{\delta}{2}$ of the half-window and uses $f(x)$ .

Suppose the definition dropped both requirements, and only asked that “ $x \in \dom f$ and $a < x < a + \delta$ imply $\abs{f(x) - L} < \eps$ ”. Take $f(x) = \sqrt{-x}$ , which is defined exactly for $x \le 0$ , and $a = 0$ . No $x$ in its domain satisfies $0 < x < \delta$ , so the condition holds for every $L$ , and every real number would be “the right-hand limit” of $\sqrt{-x}$ at 0. Our definition does not apply there: $\sqrt{-x}$ is defined on no interval $(0, r)$ . It does apply from the left, where $\lim_{x \to 0^{-}} \sqrt{-x} = 0$ , by the argument of Example 3 with $-x$ in place of $x$ .

Summary¶

$\lim_{x \to a^{+}} f(x) = L$ (Definition 1): for every $\eps > 0$ there is a $\delta > 0$ such that
$\boxed{ \begin{aligned} &a < x < a + \delta \\ &\quad \implies \abs{f(x) - L} < \eps , \end{aligned} }$
(16)
and the left-hand limit uses $a - \delta < x < a$ . The value $f(a)$ and the values on the other side play no part.
One-sided limits are unique (Remark 1).
$\lim_{x \to a} f(x) = L$ exactly when both one-sided limits are $L$ (Theorem 1). If they differ, or one does not exist, the limit does not exist (Corollary 1).
The jump of $f$ at $a$ is the right-hand limit minus the left-hand limit (Remark 2); it is 0 exactly when they are equal.
For a piecewise function, each half-window sees one formula; the value at $a$ never matters.
At an endpoint of the domain, such as 0 for $\sqrt{x}$ , only one one-sided limit is defined: $\lim_{x \to 0^{+}} \sqrt{x} = 0$ .

Exercises¶

Exercise 6 (The largest δ on each side)

Let $g(x) = x + 3$ for $x < 2$ and $g(x) = 3x - 1$ for $x \ge 2$ , and let $\eps = 0.03$ . Find the largest $\delta > 0$ such that $\abs{g(x) - 5} < 0.03$ for every $x$ (a) with $2 < x < 2 + \delta$ , (b) with $2 - \delta < x < 2$ , (c) with $0 < \abs{x - 2} < \delta$ .

Hint 1

On each side, write $\abs{g(x) - 5}$ as a multiple of $\abs{x - 2}$ .

Hint 2

For (c), the window $0 < \abs{x - 2} < \delta$ is made of the two half-windows of (a) and (b) (part (e) of the proposition on distance inequalities).

Answer

(a) $\frac{1}{100}$ (b) $\frac{3}{100}$ (c) $\frac{1}{100}$

Where this leads¶

These topics build on this one:

Infinite Limits and Vertical Asymptotes
Continuity (coming soon)

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The theorem of this page and its corollary are the usual way to decide whether a limit exists: find the two one-sided limits and compare them.