Limits

A limit asks where a function is heading as its input gets close to a point. Start by watching nearby values, not by substituting the point.

Source code

What does approaching mean?

Pick a target on the horizontal axis. Move toward it from the left and right, and watch the function’s height. For a limit to exist from both sides, those heights must head toward the same value.

\lim_{x\to a}f(x)=L\quad\Longleftrightarrow\quad\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L

The function can have a hole or a different value at the target. A limit describes nearby inputs, so that single point does not decide the answer.

Approach a point

Choose a function, then bring the sample point closer from either side.

Blue shows the function; amber marks the approaching sample. Open circles mark a missing endpoint. The horizontal axis is

x\in[-1,3]

Example

f(x)=x^2,\quad a=2

Approach from

Distance from the target

h=0.316

x=1.684,\quad f(x)=2.835

f(2)=4

\lim_{x\to 2}f(x)=4

Look from both sides

Nearby function values from the left and right
$h$	$f(a-h)$	$f(a+h)$
$1$	$1$	$9$
$0.1$	$3.61$	$4.41$
$0.01$	$3.96$	$4.04$
$0.001$	$3.996$	$4.004$

Both sides approach the same height, and the point on the curve has that height too.

Try a limit by hand

Consider the fraction below. Substitution at the target gives an undefined fraction, but factoring shows what happens nearby.

\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1\quad(x\ne1)

The simplified expression has a height approaching $2$ from either side. The original fraction is still undefined at the target, yet its limit exists:

\lim_{x\to1}\frac{x^2-1}{x-1}=2

Check your understanding

Choose an answer to see why it works. You can change your choice.

Question 1
What value does the curve approach?
$\lim_{x\to 3}x^2$
Question 2
The expression is undefined at the target. What is its limit?
$\lim_{x\to 2}\frac{x^2-4}{x-2}$
Question 3
The left side approaches one height and the right side another. What is the two-sided limit?
$\lim_{x\to1}f(x),\quad f(x)=\begin{cases}2&x<1\\3&x\ge1\end{cases}$