The Precise Definitions Of Limits

Previously, we provided intuitive definitions of limits. While these intuitive definitions can help with understanding and may suffice for practical use, they lack rigor because they rely on imprecise language such as "as close as we please" and "sufficiently close." In this section, we aim to provide more rigorous definitions.

Quick Reference

Limit Type	Formal Definition	Key Condition
Ordinary Limit ( $lim_{x \to a} f (x) = L$ )	$\forall ϵ > 0 \exists δ > 0 \forall x$ $(0 < \| x - a \| < δ ⟹ \| f (x) - L \| < ϵ)$	$f (x)$ is trapped in $(L - ϵ, L + ϵ)$ for $x \in (a - δ, a + δ) ∖ {a}$ .
Left-Hand Limit ( $lim_{x \to a^{-}} f (x) = L$ )	$\forall ϵ > 0 \exists δ > 0 \forall x$ $(a - δ < x < a ⟹ \| f (x) - L \| < ϵ)$	$f (x)$ is within $ϵ$ of $L$ when approaching $a$ strictly from the left.
Right-Hand Limit ( $lim_{x \to a^{+}} f (x) = L$ )	$\forall ϵ > 0 \exists δ > 0 \forall x$ $(a < x < a + δ ⟹ \| f (x) - L \| < ϵ)$	$f (x)$ is within $ϵ$ of $L$ when approaching $a$ strictly from the right.
Infinite Limit ( $lim_{x \to a} f (x) = + \infty$ )	$\forall K > 0 \exists δ > 0 \forall x$ $(0 < \| x - a \| < δ ⟹ f (x) > K)$	$f (x)$ grows above any horizontal bound $K$ as $x$ gets close to $a$ .
Infinite Limit ( $lim_{x \to a} f (x) = - \infty$ )	$\forall K < 0 \exists δ > 0 \forall x$ $(0 < \| x - a \| < δ ⟹ f (x) < K)$	$f (x)$ falls below any bound $K$ as $x$ gets close to $a$ .
Limit at $+ \infty$ ( $lim_{x \to + \infty} f (x) = L$ )	$\forall ϵ > 0, \exists N > 0 \forall x$ $(x > N ⟹ \| f (x) - L \| < ϵ)$	$f (x)$ stays within $ϵ$ of $L$ for all sufficiently large positive $x > N$ .
Limit at $- \infty$ ( $lim_{x \to - \infty} f (x) = L$ )	$\forall ϵ > 0 \exists N < 0 \forall x$ $(x < N ⟹ \| f (x) - L \| < ϵ)$	$f (x)$ stays within $ϵ$ of $L$ for all sufficiently large negative $x < N$ .
Linear Proof Strategy	$lim_{x \to a} (m x + b) = m a + b$	Set $δ = \frac{ϵ}{\| m \|}$ (for $m \neq 0$ ).
Nonlinear Proof Strategy	$lim_{x \to a} f (x) = L$	Bound non- $\| x - a \|$ factors locally (e.g., $δ_{1} = 1$ ), then set $δ = min {1, \frac{ϵ}{C}}$ .

History

The evolution of the limit concept

Although mathematicians intuitively applied limiting processes even before the development of calculus, without a precise definition of limit, they were not able to prove important theorems of calculus with sufficient rigor. The first person who tried to put the definition on a mathematically sound basis was the French mathematician, engineer, and physicist, Augustin-Louis Cauchy (1789–1857). Finally the definitive modern definition of limit was formulated by the German mathematician Karl Weierstrass (1815–1897) who used two Greek letters $ϵ$ (epsilon) and $δ$ (delta) for the small differences.

The $ϵ$ – $δ$ Definition of a Limit

Let's revisit the intuitive definition of $lim_{x \to a} f (x) = L$ : The limit of $f (x)$ as x approaches a is $L$ if we can make the values of $f (x)$ as close to $L$ as we please by taking x sufficiently close to a, but not equal to a.

Let's express every element in the above description in mathematical language. We say two quantities A and B are close when the distance between them is small. Because the distance from x to a is $| x - a |$ and the distance from $f (x)$ to $L$ is $| f (x) - L |$ , we can alternatively say that the limit of $f (x)$ as x approaches a is $L$ if we can make $| f (x) - L |$ "as small as we please" by making $| x - a |$ "sufficiently small," but not zero.

Now, we need to precisely define "as small as we please" and "sufficiently small" mathematically. To achieve this, imagine a game between you and me. You challenge me by giving a small number $ϵ > 0$ (as small as you wish) and my task is to find another number $δ > 0$ such that $| f (x) - L | < ϵ$ for all $x \neq a$ satisfying $| x - a | < δ$ .

Because the inequality $| f (x) - L | < ϵ$ is equivalent to

- ϵ < f (x) - L < ϵ

L - ϵ < f (x) < L + ϵ,

the geometrical meaning of this game is as follows: You consider a band of width $2 ϵ$ bounded by the lines $y = L - ϵ$ and $y = L + ϵ$ (see the following figure), and I must identify an open interval of radius $δ$ with a at the center such that all the points on the graph of $y = f (x)$ that correspond to values of x within the interval $(a - δ, a + δ)$ —except for the point directly above a—must fall within the band you specified.

Geometric interpretation of epsilon-delta limit showing horizontal band around L and vertical strip around a — Geometric interpretation of the $ϵ$ – $δ$ definition of a limit.

Let's explore, with an example, what it means when you give me an $ϵ$ , and I can find a corresponding $δ$ .

For example, previously, we saw that if

f (x) = \frac{2 x^{2} - 2}{x - 1},

then $lim_{x \to 1} f (x) = 4$ . For instance, if you give me $ϵ = 0.01$ , I will take $δ = 0.005$ (or smaller), and claim $| f (x) - 4 | < 0.01$ for all $x \neq 1$ satisfying $| x - 1 | < 0.005$ ; because if $x \neq 1$ and $| x - 1 | < 0.005$ then

\begin{aligned} |f(x) - 4| &= |2x + 2 - 4| \\ &= |2x - 2| \\ &= 2|x - 1| < 2 \times 0.005 = 0.01. \end{aligned}

Recall that when $x \neq 1$ :

f (x) = \frac{2 (x^{2} - 1)}{x - 1} = 2 \frac{(x - 1) (x + 1)}{x - 1} = 2 (x + 1) .

If you give me $ϵ = 0.0002$ , I just need to take $δ = 0.0001$ (or smaller), because $| x - 1 | < 0.0001$ and $x \neq 1$ implies that $| f (x) - 4 | < 0.0002$ :

\begin{aligned} |f(x) - 4| &= |2x + 2 - 4| \\ &= 2|x - 1| < 2 \times 0.0001 = 0.0002. \end{aligned}

If this game goes on forever and for every $ϵ > 0$ you give me, I can find a $δ > 0$ with the aforementioned conditions, then we say the limit of $f (x)$ as x approaches a is $L$ .

Specifically, we state that if we can make $| f (x) - L |$ less than any given positive number $ϵ > 0$ whenever $| x - a |$ is less than some appropriately chosen positive number $δ$ and $| x - a | \neq 0$ (because $x \neq a$ ) then

lim_{x \to a} f (x) = L .

Remark that in general the size of $δ$ depends on the value of $ϵ$ .

Instead of stating " $| x - a | < δ$ and $| x - a | \neq 0$ (or $x \neq a$ )," we can express it as $0 < | x - a | < δ$ .
Recall that the condition $x \neq a$ or $0 < | x - a |$ is imposed because what happens to $f (x)$ when x equals a has no influence on the value or the existence of the limit. The focus is solely on the behavior of $f (x)$ for values of x that are close to a.

Definition: Let f be a function that is defined at every number in some open interval containing a except possibly at the number a itself. The symbol $lim_{x \to a} f (x) = L$ means that for every $ϵ > 0$ , however small, there exists some $δ > 0$ such that

| f (x) - L | < ϵ whenever 0 < | x - a | < δ .

Another way of writing the last line is: "for all x:

0 < | x - a | < δ ⟹ | f (x) - L | < ϵ

We use the symbol " $⟹$ " in place of "implies" or "if ... then ... ." The above definition is commonly referred to as the epsilon-delta definition of a limit.

Instead of saying "let f be a function that is defined at every number in some open interval containing a except possibly at the number a itself", we can say "let f be defined in a deleted neighborhood of the point a." For the definition of the deleted neighborhood, see here.

Also instead of saying "for every $ϵ > 0$ , there exists a $δ$ such that ...", we can say for every neighborhood of $L$ , $N_{1} (L)$ , there is some neighborhood of $a$ , $N_{2} (a)$ such that

f (x) \in N_{1} (L) whenever x \in N_{2} (a) and x \neq a .

Using mathematical symbols

Mathematics uses symbols as shorthand for ideas that would otherwise take a full sentence to state. Three of them appear throughout this book:

$\forall$ , an upside-down A, stands for "for all" or "for every"
$\exists$ , a backward E, stands for "there exists"
$⟹$ stands for "implies", so $P ⟹ Q$ is read "if $P$ , then $Q$ "

With this notation, the statement $lim_{x \to a} f (x) = L$ becomes

\forall ε > 0 \exists δ > 0 \forall x \in D_{f} (0 < | x - a | < δ ⟹ | f (x) - L | < ε)

Read it aloud as: for every $ε > 0$ there exists a $δ > 0$ such that, for every $x$ in the domain of $f$ , if $0 < | x - a | < δ$ , then $| f (x) - L | < ε$ .

Let the function f be defined by $f (x) = 5 x + 1$ . Given $lim_{x \to 0} f (x) = 1$ , find a $δ$ such that

| f (x) - 1 | < 0.01 whenever 0 < | x | < δ .

Solution

Here $ϵ = 0.01$ is given. To find a $δ$ , we need to establish a connection between

| f (x) - L | = | (5 x + 1) - 1 | and | x - a | = | x | .

We notice that

| f (x) - 1 | = | 5 x + 1 - 1 | = 5 | x |

Therefore, we want

5 | x | < 0.01 whenever 0 < | x | < δ

If we take $δ = 0.01 / 5 = 0.002$ , then we have

5 | x | < 5 \times 0.002 = 0.01 whenever 0 < | x | < 0.002

Note that $0.002$ is the largest value that we can choose for $δ$ ; any number less than $0.002$ for $δ$ also works. That is, if $0 < δ < 0.002$ , then

| f (x) - 1 | < 0.01, whenever 0 < | x - 0 | < δ

because any number x that satisfies $0 < | x - 0 | < δ$ also satisfies $0 < | x - 0 | < 0.002$ .

To prove using the $ϵ$ – $δ$ definition, we suppose $ϵ > 0$ is given.

To find an appropriately chosen $δ$ , we need to establish a connection between $0 < | x - a | < δ$ and $| f (x) - L | < ϵ$ .

To this end, work backward from the inequality $| f (x) - L | < ϵ$ and simplify it until we get $| x - a | < g (ϵ)$ , where $g (ϵ)$ is some function of $ϵ$ . We then choose $δ \leq g (ϵ)$ .

Use the $ϵ$ – $δ$ definition to prove

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

Solution

Here $f (x) = 2 x - 3$ , $a = 1$ , and $L = - 1$ . Suppose $ϵ > 0$ is given. We want to find a number $δ > 0$ such that

0 < | x - 1 | < δ ⟹ | 2 x - 3 - (- 1) | < ϵ .

We start with $| 2 x - 3 + 1 | < ϵ$ and simplify it:

| 2 x - 3 + 1 | = | 2 x - 2 | = 2 | x - 1 | < ϵ

or equivalently

| x - 1 | < \frac{1}{2} ϵ .

So if we choose $δ \leq ϵ / 2$ , then

0 < | x - 1 | < δ \leq ϵ / 2 ⟹ | 2 x - 3 - (- 1) | < ϵ .

In general:

Theorem: If $f (x) = m x + b$ where m and b are any real numbers then $lim_{x \to a} f (x) = m a + b$ for any real number a.

Proof

Let $ϵ > 0$ be given. Then we need to show that there exists some $δ > 0$ such that if $0 < | x - a | < δ$ then

| f (x) - (m a + b) | = | m x + b - (m a + b) | = | m (x - a) | < ϵ .

From the above inequality, it is clear that if $δ < \frac{ϵ}{| m |}$ (provided $m \neq 0$ ), then for all x

0 < | x - a | < δ < \frac{ϵ}{| m |} ⟹ | f (x) - m a - b | < ϵ .

In the case that $m = 0$ , the inequality $| m (x - a) | = 0 < ϵ$ holds for all values of x, regardless of the choice of $δ > 0$ .

Use the $ϵ$ – $δ$ definition to prove

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

Solution

Here $f (x) = x^{2}, a = 2$ , and $L = 4$ . Suppose $ϵ > 0$ is given. We want to find a number $δ > 0$ such that

0 < | x - 2 | < δ ⟹ | x^{2} - 4 | < ϵ

To establish a connection between $| x^{2} - 4 |$ and $| x - 2 |$ , we factor $x^{2} - 4 = (x - 2) (x + 2)$ :

| x^{2} - 4 | = | (x - 2) (x + 2) | = | x - 2 | | x + 2 | .

So we want

0 < | x - 2 | < δ ⟹ | x - 2 | | x + 2 | < ϵ .

If we can find a positive constant C such that $| x + 2 | < C$ for x close to 2 and choose $δ \leq ϵ / C$ then

0 < | x - 2 | < \frac{ϵ}{C} ⟹ | x - 2 | | x + 2 | < \frac{ϵ}{C} C = ϵ .

So the question is: how can we find C? We can find such a number C if we restrict x to some neighborhood of 2 (= an interval with center at 2). For example, because x gets closer and closer to 2, we can assume that x is restricted to the numbers that are closer to 2 than 1 unit; that is $| x - 2 | < 1$ or $- 1 < x - 2 < 1$ or $1 < x < 3$ .

If we add 2 to each side and then apply the absolute value we get $3 < x + 2 < 5$ , so $3 < | x + 2 | < 5$ .

So if $| x - 2 | < 1$ then $| x + 2 | < 5$ and

| x^{2} - 4 | = | x - 2 | | x + 2 | < 5 | x - 2 |

Therefore, if we have two restrictions on x, namely

| x - 2 | < 1 and | x - 2 | < \frac{1}{5} ϵ,

then we will have

| x^{2} - 4 | < 5 | x - 2 | < ϵ .

To satisfy both inequalities $| x - 2 | < 1$ and $| x - 2 | < ϵ / 5$ , we let $δ$ be the minimum of 1 and $ϵ / 5$ :

δ \leq min {1, ϵ / 5},

and we conclude

0 < | x - 2 | < δ ⟹ | x^{2} - 4 | < ϵ .

For example, if $ϵ = 0.25$ , then we choose $min {1, 0.25 / 5} = 0.05$ (or less) for $δ$ . The following figure shows when $ϵ = 0.25$ and $δ = 0.05$ .

Epsilon-delta graph for y = x^2 with epsilon = 0.25 and delta = 0.05 — Illustration of $ϵ = 0.25$ and $δ = 0.05$ for $f (x) = x^{2}$ at $x = 2$ .

Notice that somebody else might say because x has to be sufficiently close to 2, we should concern ourselves to only those values of x that are not farther away from 2 than 0.1 units; that is, the values of x for which $| x - 2 | < 0.1$ . In this case $1.9 < x < 2.1$ ; adding 2 to each side, we get $3.9 < x + 2 < 4.1$ or $3.9 < | x + 2 | < 4.1$ . Here $4.1 \leq C$ . So if we have two restrictions on x: $| x - 2 | < 0.1$ and $| x - 2 | < ϵ / 4.1$ , then we will have $| x^{2} - 4 | < ϵ$ . Thus another choice for $δ$ is $δ \leq min {0.1, ϵ / 4.1}$ . In this case, if $ϵ = 0.25$ is given, we choose $min {0.1, 0.25 / 4.1} \approx 0.061$ (or less) for $δ$ .

Use the $ϵ$ – $δ$ definition to prove

lim_{x \to \frac{1}{2}} \frac{1}{x^{2}} = 4.

Solution

Here $f (x) = 1 / x^{2}, a = 1 / 2$ , and $L = 4$ . We need to show that for every $ϵ > 0$ , there exists $δ > 0$ such that

0 < | x - \frac{1}{2} | < δ ⟹ | \frac{1}{x^{2}} - 4 | < ϵ .

Suppose $ϵ > 0$ is given. To establish a connection between $| x - 1 / 2 |$ and $| 1 / x^{2} - 4 |$ , we work backward from $| 1 / x^{2} - 4 |$ and simplify it:

\begin{aligned} \left|\frac{1}{x^2} - 4\right| &= \left|\frac{1 - 4x^2}{x^2}\right| \\ &= \left|\frac{4(\frac{1}{4} - x^2)}{x^2}\right| \\ &= \frac{4}{x^2} \left|\frac{1}{2} - x\right| \left|\frac{1}{2} + x\right| \\ &= \frac{4}{x^2} \left|x - \frac{1}{2}\right| \left|x + \frac{1}{2}\right| \end{aligned}

Here we have factored $\frac{1}{4} - x^{2} = {(\frac{1}{2})}^{2} - x^{2}$ using the Difference of Squares formula ( $A^{2} - B^{2} = (A - B) (A + B)$ ). Now we need to find an upper bound for $\frac{4}{x^{2}} | x + 1 / 2 |$ when x is close to $1 / 2$ ; that is to find a constant $C > 0$ such that for x close to $1 / 2$ :

\frac{4}{x^{2}} | x + \frac{1}{2} | < C .

Similar to the previous example, we proceed by restricting x to lie in some neighborhood of $1 / 2$ (= an interval centered at $1 / 2$ ), but unlike the previous example, the radius of the neighborhood cannot be 1 because $\frac{4}{x^{2}} | x + 1 / 2 |$ is not defined for $x = 0$ . So we consider a neighborhood such that it does not include $x = 0$ . For example, we assume that x is within a distance $1 / 4$ from $1 / 2$ ; that is, $| x - \frac{1}{2} | < \frac{1}{4}$ , or $\frac{1}{4} < x < \frac{3}{4}$ .

When $1 / 4 < x < 3 / 4$ :

\frac{3}{4} < x + \frac{1}{2} < \frac{5}{4} and \frac{1}{16} < x^{2} < \frac{9}{16} .

Therefore in this neighborhood of $1 / 2$ :

| x + \frac{1}{2} | < \frac{5}{4}, \frac{16}{9} < \frac{1}{x^{2}} < 16

and finally

\frac{4}{x^{2}} | x + \frac{1}{2} | < 4 \times 16 \times \frac{5}{4} = 80.

This means that if $| x - 1 / 2 | < 1 / 4$ , then

| \frac{1}{x^{2}} - 4 | = \frac{4}{x^{2}} | x - \frac{1}{2} | | x + \frac{1}{2} | < 80 | x - \frac{1}{2} | .

If $| x - 1 / 2 | < ϵ / 80$ , then $80 | x - 1 / 2 | < ϵ$ .

If we take $δ \leq min {\frac{1}{4}, \frac{ϵ}{80}}$ , then whenever $| x - 1 / 2 | < δ$ , certainly $| x - 1 / 2 | < 1 / 4$ and $| \frac{1}{x^{2}} - 4 | < 80 | x - 1 / 2 |$ . Additionally because $| x - 1 / 2 | < \frac{ϵ}{80}$ , we will have $| \frac{1}{x^{2}} - 4 | < 80 | x - 1 / 2 | < ϵ$ .

So we showed that for every $ϵ > 0$ , we can take $δ \leq min {\frac{1}{4}, \frac{ϵ}{80}}$ and

| \frac{1}{x^{2}} - 4 | < ϵ whenever | x - \frac{1}{2} | < δ .

Use the $ϵ$ – $δ$ definition to prove

lim_{x \to a} \sqrt{x} = \sqrt{a}, (a > 0)

Solution

We must show that for every given $ϵ > 0$ , there exists a $δ > 0$ such that for all x:

if | x - a | < δ then | \sqrt{x} - \sqrt{a} | < ϵ

We note that we must have $δ \leq a$ , otherwise, $\sqrt{x}$ will not be defined for some x (see the following figure).

Graph of y = sqrt(x) showing domain restriction delta <= a — Graph of $y = \sqrt{x}$ . If $δ > a$ , then for some x between $a - δ$ and $0$ , $\sqrt{x}$ is not defined. Therefore, we must choose $δ \leq a$ .

To express $| \sqrt{x} - \sqrt{a} |$ in terms of $| x - a |$ , we multiply and divide $\sqrt{x} - \sqrt{a}$ by its conjugate $\sqrt{x} + \sqrt{a}$ :

\begin{aligned} |\sqrt{x} - \sqrt{a}| &= \left|(\sqrt{x} - \sqrt{a}) \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}\right| \\ &= \left|\frac{x - a}{\sqrt{x} + \sqrt{a}}\right| \\ &= \frac{1}{|\sqrt{x} + \sqrt{a}|} |x - a| \end{aligned}

Note that because $a > 0$ , and $x > 0$ (if $x < 0$ , $\sqrt{x}$ would be imaginary), $\sqrt{x} + \sqrt{a} > 0$ . For the second line of the above equation, we used the identity $(A - B) (A + B) = A^{2} - B^{2}$ , and the last line follows from the fact that $| \frac{A}{B} | = \frac{| A |}{| B |}$ ( $B \neq 0$ ).

Now we need to find an upper bound for $\frac{1}{| \sqrt{x} + \sqrt{a} |} | x - a |$ and make the upper bound less than $ϵ$ . We get an upper bound for this fraction if x is replaced by 0 in the denominator, because if x is replaced by any other number $b \neq 0$ , then $0 + \sqrt{a} < \sqrt{b} + \sqrt{a}$ . That is,

\frac{1}{| \sqrt{x} + \sqrt{a} |} | x - a | < \frac{1}{\sqrt{a}} | x - a | .

If we choose $δ$ to be less than $ϵ \sqrt{a}$ (and of course less than a), then $0 < | x - a | < δ < ϵ \sqrt{a} ⟹$

| \sqrt{x} - \sqrt{a} | = \frac{1}{| \sqrt{x} + \sqrt{a} |} | x - a | < \frac{1}{\sqrt{a}} | x - a | < \frac{1}{\sqrt{a}} ϵ \sqrt{a} = ϵ .

That is, for all x: if $0 < | x - a | < δ$ then $| \sqrt{x} - \sqrt{a} | < ϵ$ , where $δ = min {a, ϵ \sqrt{a}}$ .

The Precise Definitions of One-Sided Limits

By slightly modifying the definition of two-sided (or ordinary) limits, we obtain the rigorous definitions of one-sided limits.

Let f be defined on an open interval $(b, a)$ for some $b < a$ . We say the left-hand limit of $f (x)$ as x approaches a is L, written

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

if for every $ϵ > 0$ , there exists some $δ > 0$ such that

| f (x) - L | < ϵ whenever a - δ < x < a .

Similarly

Let f be defined on an open interval $(a, b)$ for some $b > a$ . We say the right-hand limit of $f (x)$ as x approaches a is L, written

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

if for every $ϵ > 0$ , there exists some $δ > 0$ such that

| f (x) - L | < ϵ whenever a < x < a + δ .

Precise Definition of Infinite Limits

Let's revisit the intuitive meaning of $lim_{x \to a} f (x) = + \infty$ : If we can make $f (x)$ as large as we wish by taking x sufficiently close to a (but not equal to a), then we say $f (x)$ approaches $+ \infty$ as x approaches a.

Previously, we expressed "taking x sufficiently close to a" in mathematical language using $0 < | x - a | < δ$ . Now we express "making $f (x)$ as large as we wish" mathematically by saying $f (x) > K$ for any given positive number K.

The precise definition is as follows:

Let f be a function defined in some deleted neighborhood of a (that is, f is defined for all x on both sides of a except possibly at the number a itself). We say $f (x)$ approaches $+ \infty$ as x approaches a, written

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

if for every positive real number $K > 0$ , there exists a $δ > 0$ such that

f (x) > K whenever 0 < | x - a | < δ .

This means $lim_{x \to a} f (x) = + \infty$ if for any number $K > 0$ that you give me, I can determine a $δ > 0$ such that for all x closer to a than $δ$ (except when $x = a$ ), $f (x)$ lies above the horizontal line $y = K$ . A geometric illustration is shown below.

Geometric interpretation of infinite limit showing horizontal line y = K and delta strip around a — Illustration of $lim_{x \to a} f (x) = + \infty$ : Given any horizontal line $y = K$ , we can find $δ > 0$ such that $f (x)$ lies above the line $y = K$ for all x satisfying $0 < | x - a | < δ$ .

Using mathematical symbols,

lim_{x \to a} f (x) = + \infty ⟺ \forall K > 0 \exists δ > 0 \forall x (0 < | x - a | < δ ⟹ f (x) > K)

In a similar way

lim_{x \to a} f (x) = - \infty ⟺ \forall K < 0 \exists δ > 0 \forall x (0 < | x - a | < δ ⟹ f (x) < K)

Precise Definitions of Limits at Infinity

Now let's consider limits at infinity, where x becomes arbitrarily large positive ( $x \to + \infty$ ) or arbitrarily large negative ( $x \to - \infty$ ).

Recall the intuitive definition of $lim_{x \to + \infty} f (x) = L$ : The values of $f (x)$ can be made as close to L as we please by taking x sufficiently large.

To express "taking x sufficiently large" in rigorous mathematical language, we say $x > N$ for some positive number N. Similarly, for $x \to - \infty$ , "taking x sufficiently far to the left" means $x < N$ for some negative number N.

The formal definition is as follows:

Let f be a function defined on an interval $(c, + \infty)$ for some real number c. We say the limit of $f (x)$ as x approaches $+ \infty$ is L, written

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

if for every $ϵ > 0$ , there exists a corresponding number $N > 0$ such that

| f (x) - L | < ϵ whenever x > N .

Similarly, for limits as $x \to - \infty$ :

Let f be a function defined on an interval $(- \infty, c)$ for some real number c. We say the limit of $f (x)$ as x approaches $- \infty$ is L, written

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

if for every $ϵ > 0$ , there exists a corresponding number $N < 0$ such that

| f (x) - L | < ϵ whenever x < N .

Geometrically, $lim_{x \to + \infty} f (x) = L$ means that for any horizontal strip of width $2 ϵ$ bounded by $y = L - ϵ$ and $y = L + ϵ$ , there exists a vertical line $x = N$ such that the graph of $y = f (x)$ stays entirely within the strip for all $x > N$ .

Use the precise definition of a limit at infinity to prove:

lim_{x \to + \infty} \frac{1}{x} = 0.

Solution

Here $f (x) = 1 / x$ and $L = 0$ . Let $ϵ > 0$ be given. We must find a number $N > 0$ such that:

| \frac{1}{x} - 0 | < ϵ whenever x > N .

Since $x > N > 0$ , we have $x > 0$ , so $| \frac{1}{x} | = \frac{1}{x}$ . The target inequality $\frac{1}{x} < ϵ$ is equivalent to:

x > \frac{1}{ϵ} .

Therefore, if we choose $N = \frac{1}{ϵ}$ , then whenever $x > N = \frac{1}{ϵ}$ , we have:

| \frac{1}{x} - 0 | = \frac{1}{x} < \frac{1}{N} = ϵ .

This completes the proof.

Exercises

Use the $ϵ$ – $δ$ definition to prove:

lim_{x \to 3} (4 - 3 x) = - 5.

Solution

Here $f (x) = 4 - 3 x$ , $a = 3$ , and $L = - 5$ . Let $ϵ > 0$ be given. We need to find a $δ > 0$ such that for all x:

0 < | x - 3 | < δ ⟹ | (4 - 3 x) - (- 5) | < ϵ .

We simplify the expression inside the absolute value:

| (4 - 3 x) - (- 5) | = | 9 - 3 x | = | - 3 (x - 3) | = | - 3 | | x - 3 | = 3 | x - 3 | .

We want $3 | x - 3 | < ϵ$ , which is equivalent to:

| x - 3 | < \frac{ϵ}{3} .

Therefore, if we choose $δ = \frac{ϵ}{3}$ (or any positive number $δ \leq \frac{ϵ}{3}$ ), then whenever $0 < | x - 3 | < δ$ , we have:

| (4 - 3 x) - (- 5) | = 3 | x - 3 | < 3 \times \frac{ϵ}{3} = ϵ .

This completes the proof.

For the limit $lim_{x \to 2} (x^{2} + 3 x) = 10$ , find a value of $δ > 0$ such that $| (x^{2} + 3 x) - 10 | < 0.05$ whenever $0 < | x - 2 | < δ$ .

Solution

Here $ϵ = 0.05$ . We factor the expression $| f (x) - L |$ :

| x^{2} + 3 x - 10 | = | (x - 2) (x + 5) | = | x - 2 | | x + 5 | .

We need to bound $| x + 5 |$ near $x = 2$ . Assume x is restricted to a 1-unit neighborhood around $x = 2$ ; that is, $| x - 2 | < 1$ . This implies:

- 1 < x - 2 < 1 ⟹ 1 < x < 3.

Adding 5 to each part:

6 < x + 5 < 8 ⟹ | x + 5 | < 8.

Thus, whenever $| x - 2 | < 1$ , we have:

| x^{2} + 3 x - 10 | = | x - 2 | | x + 5 | < 8 | x - 2 | .

To make $8 | x - 2 | < 0.05$ , we require $| x - 2 | < \frac{0.05}{8} = \frac{1}{160} = 0.00625$ .

To satisfy both conditions ( $| x - 2 | < 1$ and $| x - 2 | < 0.00625$ ), we set:

δ = min {1, 0.00625} = 0.00625 .

Thus, taking $δ = 0.00625$ guarantees that $| (x^{2} + 3 x) - 10 | < 0.05$ whenever $0 < | x - 2 | < δ$ .

Use the $ϵ$ – $δ$ definition to prove:

lim_{x \to 3} x^{2} = 9.

Solution

Here $f (x) = x^{2}$ , $a = 3$ , and $L = 9$ . Let $ϵ > 0$ be given. We want to find a $δ > 0$ such that:

0 < | x - 3 | < δ ⟹ | x^{2} - 9 | < ϵ .

Factoring $| x^{2} - 9 |$ gives:

| x^{2} - 9 | = | (x - 3) (x + 3) | = | x - 3 | | x + 3 | .

To find an upper bound for $| x + 3 |$ , assume x is within distance 1 of 3, so $| x - 3 | < 1$ :

- 1 < x - 3 < 1 ⟹ 2 < x < 4.

Adding 3 to each part gives $5 < x + 3 < 7$ , so $| x + 3 | < 7$ .

Therefore, if $| x - 3 | < 1$ , then:

| x^{2} - 9 | = | x - 3 | | x + 3 | < 7 | x - 3 | .

To ensure $7 | x - 3 | < ϵ$ , we require $| x - 3 | < \frac{ϵ}{7}$ .

We choose $δ = min {1, \frac{ϵ}{7}}$ . Then whenever $0 < | x - 3 | < δ$ , we have $| x - 3 | < 1$ (so $| x + 3 | < 7$ ) and $| x - 3 | < \frac{ϵ}{7}$ , giving:

| x^{2} - 9 | = | x - 3 | | x + 3 | < 7 | x - 3 | < 7 (\frac{ϵ}{7}) = ϵ .

This completes the proof.

Use the $ϵ$ – $δ$ definition to prove:

lim_{x \to 2} \frac{1}{x} = \frac{1}{2} .

Solution

Here $f (x) = 1 / x$ , $a = 2$ , and $L = 1 / 2$ . Let $ϵ > 0$ be given. We must find a $δ > 0$ such that:

0 < | x - 2 | < δ ⟹ | \frac{1}{x} - \frac{1}{2} | < ϵ .

We simplify the difference:

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

To bound $\frac{1}{2 | x |}$ , we restrict x to be within distance 1 of 2; that is, $| x - 2 | < 1$ :

- 1 < x - 2 < 1 ⟹ 1 < x < 3.

Since $x > 1$ , we have $| x | > 1$ , which implies:

\frac{1}{| x |} < 1 ⟹ \frac{1}{2 | x |} < \frac{1}{2} .

Thus, whenever $| x - 2 | < 1$ , we have:

| \frac{1}{x} - \frac{1}{2} | = \frac{| x - 2 |}{2 | x |} < \frac{1}{2} | x - 2 | .

To make $\frac{1}{2} | x - 2 | < ϵ$ , we set $| x - 2 | < 2 ϵ$ .

We choose $δ = min {1, 2 ϵ}$ . Then whenever $0 < | x - 2 | < δ$ , it follows that:

| \frac{1}{x} - \frac{1}{2} | < \frac{1}{2} | x - 2 | < \frac{1}{2} (2 ϵ) = ϵ .

This completes the proof.

Use the $ϵ$ – $δ$ definition to prove:

lim_{x \to 4} \sqrt{x} = 2.

Solution

Here $f (x) = \sqrt{x}$ , $a = 4$ , and $L = 2$ . Let $ϵ > 0$ be given. We need to find $δ > 0$ such that:

0 < | x - 4 | < δ ⟹ | \sqrt{x} - 2 | < ϵ .

Multiplying and dividing by the conjugate $\sqrt{x} + 2$ :

| \sqrt{x} - 2 | = | (\sqrt{x} - 2) \frac{\sqrt{x} + 2}{\sqrt{x} + 2} | = \frac{| x - 4 |}{\sqrt{x} + 2} .

Since $\sqrt{x} \geq 0$ for all $x \geq 0$ , we have $\sqrt{x} + 2 \geq 2$ , so:

\frac{1}{\sqrt{x} + 2} \leq \frac{1}{2} .

Thus:

| \sqrt{x} - 2 | = \frac{| x - 4 |}{\sqrt{x} + 2} \leq \frac{1}{2} | x - 4 | .

To make $\frac{1}{2} | x - 4 | < ϵ$ , we need $| x - 4 | < 2 ϵ$ .

We choose $δ = min {4, 2 ϵ}$ . Then whenever $0 < | x - 4 | < δ$ :

| \sqrt{x} - 2 | \leq \frac{1}{2} | x - 4 | < \frac{1}{2} (2 ϵ) = ϵ .

This completes the proof.

Use the $ϵ$ – $δ$ definition to prove:

lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} = - \frac{1}{4} .

Solution

Here $f (x) = \frac{\frac{1}{x} - \frac{1}{2}}{x - 2}$ , $a = 2$ , and $L = - 1 / 4$ . Let $ϵ > 0$ be given. We must find a $δ > 0$ such that for all x:

0 < | x - 2 | < δ ⟹ | \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} - (- \frac{1}{4}) | < ϵ .

First, for $x \neq 2$ , we simplify $f (x)$ :

f (x) = \frac{\frac{2 - x}{2 x}}{x - 2} = \frac{- (x - 2)}{2 x (x - 2)} = - \frac{1}{2 x} .

Now we simplify the expression inside the absolute value:

| f (x) - (- \frac{1}{4}) | = | - \frac{1}{2 x} + \frac{1}{4} | = | \frac{- 2 + x}{4 x} | = \frac{| x - 2 |}{4 | x |} .

To bound $\frac{1}{4 | x |}$ , we restrict x to be within distance 1 of 2; that is, $| x - 2 | < 1$ :

- 1 < x - 2 < 1 ⟹ 1 < x < 3.

Since $x > 1$ , we have $| x | > 1$ , which implies $\frac{1}{| x |} < 1$ and therefore:

\frac{1}{4 | x |} < \frac{1}{4} .

Thus, whenever $| x - 2 | < 1$ , we have:

| f (x) - (- \frac{1}{4}) | = \frac{| x - 2 |}{4 | x |} < \frac{1}{4} | x - 2 | .

To make $\frac{1}{4} | x - 2 | < ϵ$ , we set $| x - 2 | < 4 ϵ$ .

We choose $δ = min {1, 4 ϵ}$ . Then whenever $0 < | x - 2 | < δ$ , it follows that:

| \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} - (- \frac{1}{4}) | < \frac{1}{4} | x - 2 | < \frac{1}{4} (4 ϵ) = ϵ .

This completes the proof.

Use the $ϵ$ – $δ$ definition to prove:

lim_{x \to - 5} \frac{x^{2} + 6 x + 5}{x + 5} = - 4.

Solution

Here $f (x) = \frac{x^{2} + 6 x + 5}{x + 5}$ , $a = - 5$ , and $L = - 4$ . Let $ϵ > 0$ be given. We need to find a $δ > 0$ such that for all x:

0 < | x - (- 5) | < δ ⟹ | \frac{x^{2} + 6 x + 5}{x + 5} - (- 4) | < ϵ .

For $x \neq - 5$ , we factor the numerator:

f (x) = \frac{(x + 5) (x + 1)}{x + 5} = x + 1.

Now we substitute this into the absolute value expression:

| f (x) - (- 4) | = | (x + 1) - (- 4) | = | x + 5 | .

We want $| x + 5 | < ϵ$ . Note that $| x - (- 5) | = | x + 5 |$ .

Therefore, if we choose $δ = ϵ$ , then whenever $0 < | x - (- 5) | < δ$ , we have:

| \frac{x^{2} + 6 x + 5}{x + 5} - (- 4) | = | x + 5 | < ϵ .

This completes the proof.

Use the $ϵ$ – $δ$ definition to prove:

lim_{x \to 3} \frac{x^{3} - 27}{x - 3} = 27.

Solution

Here $f (x) = \frac{x^{3} - 27}{x - 3}$ , $a = 3$ , and $L = 27$ . Let $ϵ > 0$ be given. We want to find a $δ > 0$ such that for all x:

0 < | x - 3 | < δ ⟹ | \frac{x^{3} - 27}{x - 3} - 27 | < ϵ .

For $x \neq 3$ , we factor the difference of cubes $x^{3} - 27 = (x - 3) (x^{2} + 3 x + 9)$ :

f (x) = \frac{(x - 3) (x^{2} + 3 x + 9)}{x - 3} = x^{2} + 3 x + 9.

Substituting this into the absolute value expression gives:

| f (x) - 27 | = | (x^{2} + 3 x + 9) - 27 | = | x^{2} + 3 x - 18 | .

Factoring $| x^{2} + 3 x - 18 |$ yields:

| x^{2} + 3 x - 18 | = | (x - 3) (x + 6) | = | x - 3 | | x + 6 | .

To find an upper bound for $| x + 6 |$ , assume x is restricted to a 1-unit neighborhood of 3; that is, $| x - 3 | < 1$ :

- 1 < x - 3 < 1 ⟹ 2 < x < 4.

Adding 6 to each part gives $8 < x + 6 < 10$ , so $| x + 6 | < 10$ .

Therefore, if $| x - 3 | < 1$ , then:

| f (x) - 27 | = | x - 3 | | x + 6 | < 10 | x - 3 | .

To ensure $10 | x - 3 | < ϵ$ , we require $| x - 3 | < \frac{ϵ}{10}$ .

We choose $δ = min {1, \frac{ϵ}{10}}$ . Then whenever $0 < | x - 3 | < δ$ , we have $| x - 3 | < 1$ (so $| x + 6 | < 10$ ) and $| x - 3 | < \frac{ϵ}{10}$ , which implies:

| \frac{x^{3} - 27}{x - 3} - 27 | = | x - 3 | | x + 6 | < 10 | x - 3 | < 10 (\frac{ϵ}{10}) = ϵ .

This completes the proof.

Use the precise definition of a one-sided infinite limit to prove:

lim_{x \to 0^{+}} \ln x = - \infty .

Solution

Here $f (x) = \ln x$ and $a = 0$ . By definition, $lim_{x \to 0^{+}} \ln x = - \infty$ means that for every negative real number $M < 0$ , there exists a $δ > 0$ such that:

\ln x < M whenever 0 < x < δ .

Given $M < 0$ , we solve the inequality $\ln x < M$ for x. Since the exponential function $e^{t}$ is strictly increasing, applying it to both sides yields:

e^{\ln x} < e^{M} ⟹ x < e^{M} .

Therefore, if we choose $δ = e^{M}$ (note $e^{M} > 0$ for any real $M$ ), then whenever $0 < x < δ = e^{M}$ , taking the natural logarithm of both sides gives:

\ln x < \ln (e^{M}) = M .

This completes the proof.

Use the precise definition of an infinite limit to prove:

lim_{x \to 1} \frac{1}{(x - 1)^{2}} = + \infty .

Solution

Here $f (x) = \frac{1}{(x - 1)^{2}}$ and $a = 1$ . Let $K > 0$ be a given positive real number. We need to find a $δ > 0$ such that for all $x \neq 1$ :

\frac{1}{(x - 1)^{2}} > K whenever 0 < | x - 1 | < δ .

We rearrange the target inequality $\frac{1}{(x - 1)^{2}} > K$ :

(x - 1)^{2} < \frac{1}{K} ⟺ | x - 1 | < \frac{1}{\sqrt{K}} .

Therefore, if we choose $δ = \frac{1}{\sqrt{K}}$ , then whenever $0 < | x - 1 | < δ = \frac{1}{\sqrt{K}}$ , we square both sides to obtain:

(x - 1)^{2} < \frac{1}{K} .

Taking the reciprocal yields:

\frac{1}{(x - 1)^{2}} > K .

This completes the proof.

Use the precise definition of a limit at infinity to prove:

lim_{x \to + \infty} \frac{\sin x}{x} = 0.

Solution

Here $f (x) = \frac{\sin x}{x}$ and $L = 0$ . Let $ϵ > 0$ be given. We need to find a number $N > 0$ such that for all $x > N$ :

| \frac{\sin x}{x} - 0 | < ϵ .

Since $| \sin x | \leq 1$ for all real numbers x, whenever $x > N > 0$ , we have $x > 0$ and:

| \frac{\sin x}{x} | = \frac{| \sin x |}{x} \leq \frac{1}{x} .

To make $\frac{1}{x} < ϵ$ , we require $x > \frac{1}{ϵ}$ .

Therefore, if we choose $N = \frac{1}{ϵ}$ , then whenever $x > N = \frac{1}{ϵ}$ , we have $x > 0$ and:

| \frac{\sin x}{x} - 0 | \leq \frac{1}{x} < \frac{1}{N} = ϵ .

This completes the proof.

Use the precise definition of a limit at infinity to prove:

lim_{x \to - \infty} \arctan x = - \frac{π}{2} .

Solution

Here $f (x) = \arctan x$ and $L = - \frac{π}{2}$ . Let $ϵ > 0$ be given. Without loss of generality, assume $ϵ < π$ . We must find a number $N < 0$ such that:

| \arctan x - (- \frac{π}{2}) | < ϵ whenever x < N .

Since the range of the arctangent function is $(- \frac{π}{2}, \frac{π}{2})$ , we know that $\arctan x > - \frac{π}{2}$ for all x, so $\arctan x + \frac{π}{2} > 0$ . Thus:

| \arctan x + \frac{π}{2} | = \arctan x + \frac{π}{2} .

The target inequality $\arctan x + \frac{π}{2} < ϵ$ is equivalent to:

\arctan x < - \frac{π}{2} + ϵ .

Applying the tangent function (which is strictly increasing on $(- \frac{π}{2}, \frac{π}{2})$ ) to both sides yields:

x < \tan (- \frac{π}{2} + ϵ) = - \cot ϵ .

Therefore, if we choose $N = - \cot ϵ < 0$ , then whenever $x < N = - \cot ϵ$ , taking the arctangent of both sides gives:

\arctan x < \arctan (- \cot ϵ) = - \frac{π}{2} + ϵ .

Since $\arctan x + \frac{π}{2} > 0$ , this guarantees:

| \arctan x - (- \frac{π}{2}) | = \arctan x + \frac{π}{2} < ϵ .

This completes the proof.

Quick Reference

History

The $ϵ$ – $δ$ Definition of a Limit

Using mathematical symbols

The Precise Definitions of One-Sided Limits

Precise Definition of Infinite Limits

Precise Definitions of Limits at Infinity

Exercises

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The Precise Definitions Of Limits

Quick Reference

History

The ϵ – δ Definition of a Limit

Using mathematical symbols

The Precise Definitions of One-Sided Limits

Precise Definition of Infinite Limits

Precise Definitions of Limits at Infinity

Exercises

Quick Links

Social Media

The $ϵ$ – $δ$ Definition of a Limit