How do I resrict delta here?

Question

Prove that $\lim_{x\to 1}(x^2-1)=0$

$|(x^2 -1) -0| < \epsilon$ \* $|x^2 - 1| < \epsilon$ \* $|(x+1)(x-1)| < \epsilon$ \* $|x+1||x-1| < \epsilon$

I know I need to restrict delta now but how do I do that?

Answer 1

Hint: For $x$ sufficiently close to $1$, we have the bound $|x + 1| < 2$. Then you can estimate

$$|x^2 - 1| < 2 |x - 1| < 2 \delta$$

Answer 2

Put a bound on $$|x + 1|. $$

Use $$|x - 1| < \delta \iff 1 -\delta < x < \delta + 1.$$

Pick your favorite (positive) number as $\delta$, say $\delta = 1$, then $$x+ 1 < \delta + 1 +1 = 3.$$

Thus choose $\delta = \min \{1, \frac{\epsilon}{3} \}$ to finish the proof.

Answer 3

how $\mid x-1\mid<\delta$ then, $2 -\delta < x+1 < \delta + 2$, then $-2 -\delta < x+1 < \delta + 2$, and $\mid x+1\mid<\delta+2$ $$\mid x-1\mid\mid x+1\mid<\delta(\delta+2) $$ $$\mid x-1\mid\mid x+1\mid<(\delta+1)^2-1 $$ $$\mid x^2-1\mid<(\delta+1)^2-1 $$ you can make $$\delta=\sqrt{\epsilon+1}-1$$ for any $\epsilon$.