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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?

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

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    $\begingroup$ Always simplify the expression first! $\endgroup$ – Zhanxiong Aug 11 '15 at 1:09
  • $\begingroup$ $\frac{x^2-1}{x-1} = 1+x$ $\endgroup$ – Oussama Boussif Aug 11 '15 at 1:10
  • $\begingroup$ you can remark that $\frac{x^2-1}{x-1}=x+1$ if $x\neq 1$ $\endgroup$ – Hamza Aug 11 '15 at 1:10
  • $\begingroup$ 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 $x\neq 1$. $\endgroup$ – Amitesh Datta Aug 11 '15 at 1:11
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    $\begingroup$ Hint: $x^2 -1 $ is the difference of two squares, $\endgroup$ – Rob Arthan Aug 11 '15 at 1:11
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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 $2$.

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  • $\begingroup$ I'm kicking myself right now.... so simple. Thanks. $\endgroup$ – AJFarmar Aug 11 '15 at 1:11
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Given $\epsilon > 0$. Then $\exists \delta = \epsilon$ such that

$$|x-1| < \delta \implies |x+1 -2| = |x-1| < \delta = \epsilon$$

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