Can a limit exist if the function is undefined at that point?

Yes. A limit only depends on the outputs near the point, not at it. For example, (x^2-1)/(x-1) is undefined at x=1, but its limit there is 2.

What does it mean when a limit does not exist?

It means the outputs don't settle on one number as x approaches a. The usual reasons are that the left and right sides approach different numbers, the outputs grow without bound, or they keep oscillating.

Is the limit the same as plugging in the number?

Only for well-behaved (continuous) functions such as polynomials. Plugging in gives the value at the point; the limit describes the outputs near it. The two can differ.

What is the difference between a limit and a one-sided limit?

A one-sided limit watches the inputs from one side only: x→ a^- from the left, x→ a^+ from the right. The ordinary limit exists only when both one-sided limits exist and agree.

Why are limits important in calculus?

Derivatives and integrals are both defined as limits. The derivative is the limit of average rates of change over shorter and shorter intervals; the integral is the limit of sums of thinner and thinner rectangles.

What Is a Limit in Calculus? Explained in Plain Language

Why calculus needs limits

Calculus starts with a question that sounds simple: how fast is something changing at one exact instant? A speedometer answers it every second, but working it out by hand runs into a problem. Speed is distance divided by time, and at a single instant no time passes, so the calculation becomes $\frac{0}{0}$ , which has no value.

The way around this is to stop asking what happens at the instant and ask what happens as we get close to it. Measure the average speed over a short stretch of time, then a shorter one, then a shorter one still, and watch where the answers settle. The number they settle on is a limit, and it is the idea every other part of calculus is built on: derivatives are limits, and integrals are limits.

What does a limit mean?

Definition: limit

The limit of $f (x)$ as $x$ approaches $a$ is the number $L$ that the outputs $f (x)$ get closer and closer to as the inputs $x$ get closer and closer to $a$ (from both sides), without ever being equal to $a$ .

Two words in that definition do the work. Approaches means the inputs get as close to $a$ as we like. Without being equal to $a$ means the value of $f$ at $a$ itself plays no part: the limit only looks at the inputs near $a$ .

How do you read limit notation?

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

$lim$ is short for limit.
$x \to a$ , written underneath, says which way the input is moving: $x$ approaches $a$ .
$f (x)$ is the function whose outputs we are watching.
$= L$ says the number the outputs approach.

So $lim_{x \to 2} (3 x + 1) =$ $7$ is read aloud as "the limit of $3 x + 1$ as $x$ approaches $2$ is $7$ ".

An example: a function with a hole

Here is a function that shows why the value at the point doesn't matter. Let $f (x) =$ $\frac{x^{2} - 1}{x - 1}$ . At $x = 1$ the bottom is $0$ , so $f (1)$ is undefined. But the outputs near $1$ behave perfectly well:

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

From both sides the outputs close in on $2$ , so $lim_{x \to 1} f (x) =$ $2$ . Algebra confirms it: for $x \neq 1$ the top factors as $(x - 1) (x + 1)$ , and cancelling the shared factor $x - 1$ leaves $x + 1$ , which is close to $2$ when $x$ is close to $1$ .

The graph of

y = x + 1

with the point at

x = 1

removed, and the value moved to

f (1) = 4

. The outputs still head to

2

, so the limit is

2

even though

f (1) = 4

The graph makes the same point even more strongly. If we redefine the function so that $f (1) = 4$ , the dot moves, but every output near $x = 1$ is unchanged, so the limit is still $2$ .

Common mistake

Thinking the limit is just $f (a)$ . Often it is (for polynomials, plugging in works), but the limit and the value are different questions. A limit can exist when $f (a)$ is undefined, and the two can disagree when they both exist.

What are one-sided limits?

Sometimes the outputs approach one number from the left and a different number from the right. To describe each side we use one-sided limits: $lim_{x \to a^{-}} f (x)$ for inputs below $a$ (the left-hand limit), and $lim_{x \to a^{+}} f (x)$ for inputs above $a$ (the right-hand limit).

A function with a jump at

x = 2

: from the left the outputs approach

3

, from the right they approach

1

When does a limit exist?

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

L

exactly when both one-sided limits exist and both equal

L

Why it works

The two-sided limit asks the outputs to approach $L$ for every input close to $a$ , on either side.
If the left side heads to one number and the right side to another, no single $L$ works for both, so there is no limit.

Worked example: Reading a limit from a graph

For the jump function in the graph above, find $lim_{x \to 2^{-}} f (x)$ , $lim_{x \to 2^{+}} f (x)$ , and $lim_{x \to 2} f (x)$ .

Left of $2$ , the outputs follow the left piece up toward the open circle at height $3$ , so $lim_{x \to 2^{-}} f (x) =$ $3$ .
Right of $2$ , they follow the right piece down toward height $1$ , so $lim_{x \to 2^{+}} f (x) =$ $1$ .
The two sides disagree ( $3 \neq 1$ ), so the two-sided limit doesn't exist.

Answer: $3$ , $1$ , and the limit does not exist.

How do you find a limit?

Tables and graphs are good for seeing a limit, but they only suggest it: a table can't check every input. To find a limit exactly, try these in order:

Plug the number in. If you get an ordinary number, that's the limit. This works for polynomials and for most functions built from them, wherever they're defined.
If you get $\frac{0}{0}$ , simplify. Factor and cancel, multiply by a conjugate, or combine fractions, then plug in again.
If you get a nonzero number over $0$ , the outputs grow without bound; check the signs on each side to see whether they head to $\infty$ or $- \infty$ .

Worked example: Factor and cancel

Find $lim_{x \to 3} \frac{x^{2} - 9}{x - 3}$ .

Plugging in gives $\frac{9 - 9}{3 - 3} =$ $\frac{0}{0}$ , which doesn't decide the limit.
Factor the top: $x^{2} - 9 = (x - 3) (x + 3)$ .
For $x \neq 3$ , cancel the shared factor: $\frac{(x - 3) (x + 3)}{x - 3} =$ $x + 3$ .
Plug in again: $3 + 3 = 6$ .

Answer: $6$

Tip

The full list of methods, and how to tell which one a problem needs, is in How to evaluate limits.

What is a limit in calculus? A plain-language explanation