# Absolute value limits

How would I calculate the limit

$$\lim_{x \to 1} \frac{|x^2-1|}{x-1}?$$

I really have no idea.

I know that $$|x^2 - 1| = \begin{cases} x^2 - 1 & \text{if x \leq -1 or x \geq 1}\\ 1 - x^2 & \text{if -1 < x < 1} \end{cases}$$

but beyond this I am confused. Thanks in advance!

• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Oct 10 '18 at 0:28

You got to half of the solution yourself. As you know,

$$|x^2 - 1| = \begin{cases} x^2 - 1 & \text{if x \leq -1 or x \geq 1}\\ 1 - x^2 & \text{if -1 < x < 1} \end{cases}$$

Also to know the value of a limit, first we decide whether it exists or not. For that we must be sure that

$$\lim_{x\to a^{+}}$$ = $$\lim_{x\to a^{-}}$$

So we check the above statement:

$$\lim_{x\to 1^{+}}\frac{|x^{2}-1|}{x-1}=\lim_{x\to 1^{+}}\frac{(x-1)(x+1)}{x-1}==\lim_{x\to1^{+}}(x+1)=2$$

and

Here we use $$(1 -x ^2)$$

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

Since those two limits are not equal, the limit

$$\lim_{x \to 1} \frac{|x^2-1|}{x-1}$$

doesn't exist.

Using what you wrote, we conclude that

$$\lim_{x\to 1^{+}}\frac{|x^{2}-1|}{x-1}=\lim_{x\to 1^{+}}\frac{(x-1)(x+1)}{x-1}==\lim_{x\to1^{+}}(x+1)=2$$

and

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

Since the two one-sided limits don't agree, it follows that the limit

$$\lim_{x \to 1} \frac{|x^2-1|}{x-1}$$

doesn't exist.

Hint: Calculate The upper and lower limits separately. That is, calculate the limits as $${x\to1^+}$$ and $$x\to1^-$$ separately. This allows you to determine what the absolute value will be in each case.