What Is a Limit? One-Sided Limits and Limits from Graphs

Quick answer

The limit of f(x) as x approaches a is the value f(x) gets arbitrarily close to when x is close enough to a, without being equal to a. It describes the approach, not the arrival, so f(a) can be different or not exist at all. A limit can be estimated from a table or read from a graph. It exists only when the left-hand and right-hand limits exist and agree; jumps, unbounded growth and endless oscillation are the three ways it fails.

What you'll learn

Explain what lim f(x) = L means as x approaches a, and why f(a) does not enter into it
Estimate a limit from a table of values and read limits from a graph
Find one-sided limits and decide whether a two-sided limit exists
Recognize the three ways a limit can fail to exist

A function with a hole

The function

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

is undefined at $x = 1$ , because the denominator is $0$ there. But it is perfectly well behaved near $1$ . Put in values close to $1$ and watch the outputs:

$x$	$0.9$	$0.99$	$0.999$	$1.001$	$1.01$	$1.1$
$f(x)$	$1.9$	$1.99$	$1.999$	$2.001$	$2.01$	$2.1$

From both sides the outputs close in on $2$ . The reason shows when the numerator factors: for every $x \ne 1$ ,

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

So the graph is the line $y = x + 1$ with a single point missing.

f(x) = (x² − 1)/(x − 1)

A line with a hole at x = 1

We say the limit of $f(x)$ as $x$ approaches $1$ is $2$ , and write

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2

What the notation means

$\displaystyle\lim_{x \to a} f(x) = L$ means: $f(x)$ can be made as close to $L$ as we like by taking $x$ close enough to $a$ , with $x \ne a$ .

Two things in that sentence matter. The closeness is unlimited — within $0.001$ , within $10^{-9}$ , within any tolerance you name. And $x = a$ itself is excluded, so the value $f(a)$ plays no part.

Why the value at a does not matter

A limit answers a question about approach: where are the outputs heading as the inputs home in on $a$ ? What happens exactly at $a$ is a separate question.

That separation is the whole point. The derivative, the next big idea in calculus, is the limit of a fraction whose denominator is $0$ at the very point you care about — exactly like $f$ above. If limits depended on the value at $a$ , they could never handle those cases, and they are the cases calculus is built on.

So a function may be undefined at $a$ , or defined with some unrelated value, and the limit is unaffected either way. The limit depends only on values near $a$ , never on the value at $a$ .

Example 1 — Estimate from a table

Estimate $\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$ , with $x$ in radians.

The fraction is undefined at $0$ , so compute it nearby:

$x$	$\pm 0.1$	$\pm 0.01$	$\pm 0.001$
$\tfrac{\sin x}{x}$	$0.99833$	$0.99998$	$0.9999998$

The function is even, so both sides give the same values, and they approach $1$ . The limit is $1$ — a fact the next lesson proves, since a table suggests a limit but cannot establish one.

One-sided limits

Sometimes the outputs head to different places from the two sides. The left-hand limit uses only $x < a$ and the right-hand limit only $x > a$ :

\lim_{x \to a^-} f(x) \qquad \lim_{x \to a^+} f(x)

y = x (x < 2)
y = x − 3 (x ≥ 2)

A jump at x = 2

Here the left-hand limit at $2$ is $2$ and the right-hand limit is $-1$ . The two-sided limit asks for a single number both sides approach, and there is none.

The two-sided limit exists exactly when both one-sided limits exist and are equal, and then it is their common value.

How a limit fails to exist

There are three ways, and each has a recognizable shape on a graph.

A jump: the one-sided limits exist but differ, as at $x = 2$ above.
Unbounded growth: the values grow without bound, as $\tfrac{1}{x^2}$ does near $0$ . We write $\lim_{x \to 0} \tfrac{1}{x^2} = \infty$ to say how the limit fails; $\infty$ is not a number the values approach.
Oscillation: the values keep swinging without settling, as $\sin\!\left(\tfrac{1}{x}\right)$ does near $0$ , passing through every value from $-1$ to $1$ infinitely often.

Worked examples

Common mistakes

Practice problems

Estimate $\lim_{x \to 2} \tfrac{x^2 - 4}{x - 2}$ with a table, then confirm it by factoring.

Answer

$4$

Full solution

At $x = 1.99$ the value is $3.99$ ; at $x = 2.01$ it is $4.01$ . The values approach $4$ .

For $x \ne 2$ : $\tfrac{x^2 - 4}{x - 2} = \tfrac{(x - 2)(x + 2)}{x - 2} = x + 2$ , which approaches $4$ .
Use the graph to find $\lim_{x \to 1^-} h(x)$ , $\lim_{x \to 1^+} h(x)$ , $\lim_{x \to 1} h(x)$ and $h(1)$ .
- y = x² (x < 1)
- y = 3 − 2x (x > 1)
The function h
Answer

$1$ , $1$ , $1$ and $3$

Full solution

From the left, along $y = x^2$ , the values approach $1$ . From the right, along $y = 3 - 2x$ , they also approach $1$ .

The one-sided limits agree, so $\lim_{x \to 1} h(x) = 1$ . The filled dot gives $h(1) = 3$ , which does not affect the limit.
For $f(x) = \begin{cases} x + 1 & x < 0 \\ x^2 - 1 & x \ge 0 \end{cases}$ , find both one-sided limits at $0$ . Does $\lim_{x \to 0} f(x)$ exist?

Answer

Left: $1$ . Right: $-1$ . The limit does not exist.

Full solution

For $x < 0$ , $f(x) = x + 1 \to 1$ . For $x > 0$ , $f(x) = x^2 - 1 \to -1$ .

The one-sided limits differ, so the two-sided limit does not exist.
Describe $\lim_{x \to 3} \tfrac{1}{(x - 3)^2}$ .

Answer

It does not exist; the values grow without bound, so we write $\infty$ .

Full solution

The denominator is a small positive number on both sides of $3$ , so the fraction is large and positive and grows without bound: $\lim_{x \to 3} \tfrac{1}{(x - 3)^2} = \infty$ .
Describe $\lim_{x \to 3^-} \tfrac{1}{x - 3}$ and $\lim_{x \to 3^+} \tfrac{1}{x - 3}$ .

Answer

$-\infty$ and $\infty$

Full solution

For $x$ slightly less than $3$ the denominator is a small negative number, so the fraction is large and negative. For $x$ slightly more than $3$ it is a small positive number, so the fraction is large and positive.
A function has $\lim_{x \to 5} f(x) = 7$ . Must $f(5) = 7$ ? Must $f(5)$ exist?

Answer

Neither.

Full solution

The limit uses only values near $5$ . The function could have a hole at $5$ , or a value such as $f(5) = 0$ , and the limit would still be $7$ .
Estimate $\lim_{x \to 0} \tfrac{e^x - 1}{x}$ from values at $x = \pm 0.001$ .

Answer

$1$

Full solution

At $x = 0.001$ the value is about $1.0005$ ; at $x = -0.001$ it is about $0.9995$ . Both approach $1$ .
Which of the three ways of failing describes $\lim_{x \to 0} \cos\!\left(\tfrac{1}{x}\right)$ ?

Answer

Oscillation

Full solution

As $x \to 0$ , $\tfrac{1}{x}$ runs off to $\pm\infty$ , and cosine keeps cycling between $-1$ and $1$ . The values never settle, so the limit does not exist.
Sketch a function with $\lim_{x \to 2} f(x) = 4$ and $f(2) = 0$ .

Answer

One example: $f(x) = 2x$ for $x \ne 2$ , with $f(2) = 0$ .

Full solution

The line $y = 2x$ passes through $(2, 4)$ . Remove that point (an open circle) and place a filled dot at $(2, 0)$ .

Near $2$ the values approach $4$ , while the value at $2$ is $0$ .
A student computes $\sin\!\left(\tfrac{\pi}{x}\right)$ at $x = 0.1, 0.01, 0.001$ , gets $0$ every time, and concludes the limit at $0$ is $0$ . What went wrong?

Hint

Try $x = \tfrac{2}{5}$ or $x = \tfrac{2}{401}$ .

Answer

The sample points were special. At other points near $0$ the value is $1$ , so the limit does not exist.

Full solution

At $x = \tfrac{1}{n}$ , $\tfrac{\pi}{x} = n\pi$ and $\sin(n\pi) = 0$ — every sample the student chose.

But at $x = \tfrac{2}{4k + 1}$ , $\tfrac{\pi}{x} = \tfrac{(4k + 1)\pi}{2}$ and the sine is $1$ . These points also crowd in on $0$ .

Values $0$ and $1$ both occur arbitrarily close to $0$ , so the function oscillates and the limit does not exist.

Frequently asked questions

What is a limit in calculus?

The value f(x) approaches as x approaches a number a. It is written lim as x→a of f(x) = L, and it depends only on values of f near a, never on f(a) itself.

Can a limit exist where the function is undefined?

Yes. (x² − 1)/(x − 1) is undefined at x = 1, but its values approach 2 as x approaches 1, so the limit is 2.

When does a limit not exist?

When the left-hand and right-hand limits differ (a jump), when the values grow without bound (a vertical asymptote), or when they oscillate forever without settling.

What is a one-sided limit?

A limit taken from only one side: x → a⁻ uses values of x less than a, and x → a⁺ uses values greater than a.

Is the limit the same as f(a)?

Not in general. They agree exactly when f is continuous at a, which is the subject of the continuity lesson.

What to learn next

Limit Laws and Evaluating Limits AlgebraicallyEvaluate limits exactly: the limit laws, direct substitution, factoring and conjugates for 0/0 forms, the squeeze theorem, and why sin x over x approaches 1.

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