# Proving that the limit of $\frac{x^{2}-1}{x-1}$ as $x \rightarrow 1$ is $2$

I am trying to prove:

$$\lim _{ x \rightarrow 1 }{ \frac { x^{ 2 }-1 }{ x-1 } } = 2$$

So I began to work on proving it using epsilon-delta:

$$\left| \frac { x^{ 2 }-1 }{ x-1 } -2 \right| <\epsilon \\ -\epsilon <\frac { x^{ 2 }-1 }{ x-1 } -2<\epsilon \\ -\epsilon +2<\frac { x^{ 2 }-1 }{ x-1 } <\epsilon +2$$

And then I'm stuck. I tried reducing the with a conjugate, but that gets me nowhere.

How can I continue with this so as to reach something of this form?

$$|x - 1| <f(\epsilon)$$

• Always simplify the expression first! – Zhanxiong Aug 11 '15 at 1:09
• $\frac{x^2-1}{x-1} = 1+x$ – Oussama Boussif Aug 11 '15 at 1:10
• you can remark that $\frac{x^2-1}{x-1}=x+1$ if $x\neq 1$ – Hamza Aug 11 '15 at 1:10
• If $x\neq 1$, then $\frac{x^2-1}{x-1} = x+1$. All that matters when evaluating the limit $\lim_{x\to 1} \frac{x^2-1}{x-1}$ (and writing down an $\epsilon-\delta$ proof) are the values of the function $\frac{x^2-1}{x-1}$ when $x\neq 1$. – Amitesh Datta Aug 11 '15 at 1:11
• Hint: $x^2 -1$ is the difference of two squares, – Rob Arthan Aug 11 '15 at 1:11

Note that for $x \neq 1$ $$\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1$$ so the limit is clearly $2$.
Given $\epsilon > 0$. Then $\exists \delta = \epsilon$ such that
$$|x-1| < \delta \implies |x+1 -2| = |x-1| < \delta = \epsilon$$