Is 0/0 the answer to a limit?

No. frac00 only tells you that substituting didn't settle the limit. Simplify the expression and substitute again; the limit can turn out to be any number, or not exist.

Why can you cancel a factor that equals zero at the point?

Because a limit looks at inputs close to the number, never at the number itself. For every input the limit uses, the cancelled factor is not zero, so cancelling is valid.

What is the difference between 0/0 and a number over 0?

With frac00, both top and bottom shrink and the limit can be anything, so you simplify. With a nonzero number over 0, only the bottom shrinks, so the outputs grow without bound and the limit is ∞, -∞, or doesn't exist.

When can you use direct substitution?

Whenever the function is continuous at the number: polynomials everywhere, and fractions, roots, sin, cos, e^x and ln x wherever they're defined. If substituting gives an ordinary number, that's the limit.

Can I use L'Hôpital's rule instead of algebra?

Often, for frac00 and frac∞∞ forms, but it needs derivatives, which come later in the course. Algebra is usually quicker for polynomials and roots, and many classes expect it in the limits unit.

How to Evaluate Limits: Every Method, Step by Step

What is the first step in evaluating a limit?

The first step is always to substitute: put the number $x$ is approaching into the expression, as if you were finding the function's value. This is called direct substitution, and it works whenever the function is continuous at that number, which means its graph has no break, jump or hole there. Polynomials are continuous everywhere, and fractions, roots, $\sin$ and $\cos$ are continuous wherever they're defined, so for most limits you meet, plugging in gives the answer immediately.

Worked example: Direct substitution

Find $lim_{x \to 2} (x^{3} -$ $2 x + 5)$ .

This is a polynomial, so it is continuous everywhere and we can substitute.
$2^{3} - 2 (2) + 5 = 8 - 4 + 5$ .
$= 9$ .

Answer: $9$

What you get when you substitute tells you what to do next. There are three possible outcomes:

You get	What it means	What to do
An ordinary number, like $9$ or $\frac{10}{3}$	The function is continuous there	That number is the limit
$\frac{0}{0}$	Top and bottom share a factor that is $0$ at that number	Simplify, then substitute again
A nonzero number over $0$ , like $\frac{3}{0}$	The outputs grow without bound	Check the sign on each side

What do you do when a limit gives 0/0?

When substituting gives $\frac{0}{0}$ , the limit isn't decided yet. The $\frac{0}{0}$ form (called an indeterminate form, because it doesn't determine the answer) happens because the top and the bottom both contain a piece that equals $0$ at that number. The job is to find that shared piece and cancel it. Cancelling is allowed because, in a limit, $x$ gets close to the number but is never equal to it, so the piece you cancel is never actually $0$ .

Which algebra to use depends on what the expression looks like:

Fractions of polynomials: factor and cancel

Worked example: Factor and cancel

Find $lim_{x \to - 2} \frac{x^{2} + 5 x + 6}{x + 2}$ .

Substituting gives $\frac{4 - 10 + 6}{- 2 + 2} =$ $\frac{0}{0}$ , so simplify.
Factor the top: we need two numbers that multiply to $6$ and add to $5$ , which are $2$ and $3$ . So $x^{2} + 5 x + 6 = (x + 2) (x + 3)$ .
Cancel the shared factor $x + 2$ : $\frac{(x + 2) (x + 3)}{x + 2} =$ $x + 3$ for $x \neq - 2$ .
Substitute again: $- 2 + 3 = 1$ .

Answer: $1$

Square roots: multiply by the conjugate

When there is a square root, factoring usually isn't possible. Instead, multiply the top and the bottom by the conjugate: the same two terms with the sign between them flipped, so the conjugate of $\sqrt{x + 4} - 2$ is $\sqrt{x + 4} + 2$ . The product $(A - B) (A + B) = A^{2} - B^{2}$ squares the root away.

Worked example: Multiply by the conjugate

Find $lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$ .

Substituting gives $\frac{2 - 2}{0} =$ $\frac{0}{0}$ .
Multiply top and bottom by $\sqrt{x + 4} + 2$ . The top becomes $(\sqrt{x + 4})^{2} - 2^{2} =$ $(x + 4) - 4 =$ $x$ .
So the expression is $\frac{x}{x (\sqrt{x + 4} + 2)}$ , and cancelling $x$ leaves $\frac{1}{\sqrt{x + 4} + 2}$ .
Substitute again: $\frac{1}{\sqrt{4} + 2} =$ $\frac{1}{2 + 2} =$ $\frac{1}{4}$ .

Answer: $\frac{1}{4}$

A fraction inside a fraction: combine first

Worked example: Combine the fractions

Find $lim_{x \to 3} \frac{\frac{1}{x} - \frac{1}{3}}{x - 3}$ .

Substituting gives $\frac{\frac{1}{3} - \frac{1}{3}}{0} =$ $\frac{0}{0}$ .
Combine the top over the common bottom $3 x$ : $\frac{1}{x} - \frac{1}{3} =$ $\frac{3}{3 x} - \frac{x}{3 x} =$ $\frac{3 - x}{3 x}$ .
Notice $3 - x = - (x - 3)$ , so the top is $\frac{- (x - 3)}{3 x}$ .
Dividing by $x - 3$ cancels it: $\frac{- (x - 3)}{3 x} \cdot \frac{1}{x - 3} =$ $\frac{- 1}{3 x}$ .
Substitute again: $\frac{- 1}{3 \cdot 3} =$ $- \frac{1}{9}$ .

Answer: $- \frac{1}{9}$

Tip

If $\frac{0}{0}$ comes from a bracket raised to a power, like $\frac{(2 + h)^{2} - 4}{h}$ , multiply the bracket out first; every term left on top will then have a factor $h$ to cancel. This is the pattern in every derivative computed from the definition: see What is a derivative?

Common mistake

Cancelling terms instead of factors. In $\frac{x^{2} + 5 x + 6}{x + 2}$ you can't cross out the $x$ 's or the $2$ 's: only a factor that multiplies the whole top and the whole bottom can be cancelled. Factor first, then cancel.

What does a nonzero number over 0 mean?

If substituting gives something like $\frac{3}{0}$ , the top is heading to a number that isn't $0$ while the bottom shrinks toward $0$ . Dividing by smaller and smaller numbers makes the result bigger and bigger, so the outputs grow without bound: there is no finite limit. To say which way they grow, look at the signs just to the left and just to the right of the number.

Worked example: Checking signs on each side

Find $lim_{x \to 2} \frac{x + 1}{x - 2}$ .

Substituting gives $\frac{3}{0}$ : a nonzero number over $0$ .
Just right of $2$ (say $x = 2.01$ ): the top is about $3$ (positive) and the bottom is a tiny positive number, so the fraction is a huge positive number. $lim_{x \to 2^{+}} =$ $+ \infty$ .
Just left of $2$ (say $x = 1.99$ ): the top is still about $3$ , but the bottom is a tiny negative number, so the fraction is a huge negative number. $lim_{x \to 2^{-}} =$ $- \infty$ .
The two sides disagree, so the two-sided limit doesn't exist. The line $x = 2$ is a vertical asymptote: a vertical line the graph gets closer and closer to.

Answer: The limit does not exist ( $+ \infty$ from the right, $- \infty$ from the left).

y = \frac{x + 1}{x - 2}

near

x = 2

: the outputs shoot up on the right of the dashed line

x = 2

and down on the left.

How do you find limits with absolute values or piecewise functions?

An absolute value $| u |$ equals $u$ when $u \geq 0$ and $- u$ when $u < 0$ , so its formula changes at the point where $u = 0$ . A piecewise function also uses different formulas on different sides. In both cases, find the left-hand limit (inputs just below the number) and the right-hand limit (inputs just above) separately. The limit exists only if they agree.

Worked example: An absolute value

Find $lim_{x \to 1} \frac{| x - 1 |}{x - 1}$ .

For $x > 1$ : $x - 1 > 0$ , so $| x - 1 | = x - 1$ and the fraction is $\frac{x - 1}{x - 1} = 1$ . The right-hand limit is $1$ .
For $x < 1$ : $x - 1 < 0$ , so $| x - 1 | = - (x - 1)$ and the fraction is $\frac{- (x - 1)}{x - 1} =$ $- 1$ . The left-hand limit is $- 1$ .
$1 \neq - 1$ , so the limit doesn't exist.

Answer: The limit does not exist.

How do you evaluate limits with sine and cosine?

Trigonometric limits that give $\frac{0}{0}$ usually rest on one fact, proved in the course with the squeeze theorem: as $u \to 0$ (with $u$ in radians), $\frac{\sin u}{u} \to 1$ . Rewrite the expression so that each sine is divided by exactly what's inside it.

lim_{u \to 0} \frac{\sin u}{u} = 1

Worked example: Using the sine limit

Find $lim_{x \to 0} \frac{\sin 5 x}{3 x}$ .

Substituting gives $\frac{\sin 0}{0} =$ $\frac{0}{0}$ .
We want $5 x$ on the bottom to match the inside of the sine. Multiply and divide by $5$ : $\frac{\sin 5 x}{3 x} =$ $\frac{5}{3} \cdot \frac{\sin 5 x}{5 x}$ .
As $x \to 0$ , also $5 x \to 0$ , so $\frac{\sin 5 x}{5 x} \to 1$ .
The limit is $\frac{5}{3} \cdot 1 =$ $\frac{5}{3}$ .

Answer: $\frac{5}{3}$

When a factor like $\sin \frac{1}{x}$ wiggles forever without settling, trap the expression between two simpler ones that head to the same number; that's the squeeze theorem, covered in Lesson 1.9.

How do you choose the right method?

Substitute first, then let the result and the shape of the expression choose the method:

After substituting you see	The expression contains	Method
An ordinary number	anything	That's the limit
$\frac{0}{0}$	only polynomials	Factor and cancel
$\frac{0}{0}$	a square root	Multiply by the conjugate
$\frac{0}{0}$	a fraction inside a fraction	Combine the fractions
$\frac{0}{0}$	a bracket to a power, $h \to 0$	Multiply out, cancel $h$
$\frac{0}{0}$	$\sin$ or $\tan$ of something heading to $0$	Use $\frac{\sin u}{u} \to 1$
nonzero $\div 0$	anything	Check signs on each side
a formula that changes at the number	an absolute value or a piecewise rule	Find each one-sided limit

If $x$ is heading to $\infty$ instead of a number, the methods are different: see limits at infinity. And if algebra can't remove a $\frac{0}{0}$ (for example with $e^{x}$ or $\ln x$ ), L'Hôpital's rule usually can. The course's Playbook walks you to the right method one question at a time.

How to evaluate limits: every method, step by step