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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!

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

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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.

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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.

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