# Prove with Cauchy's limit definition ($\epsilon, \delta$) that $\lim_{x \rightarrow 0} \frac{x^2-8}{x-8}=1$

Prove with Cauchy's limit definition ($\epsilon, \delta$) that $$\lim_{x \rightarrow 0} \frac{x^2-8}{x-8}=1$$

Got really troubled with the proper technique of solving this.

Any assistance will be much appreciated!

For any $\epsilon > 0$, choose $\delta = \text{min}\left(1,\frac{7\epsilon}{2}\right)$, then: if $0 < |x| < \delta$ then $\left|\dfrac{x^2-8}{x-8} - 1\right| = \left|\dfrac{x^2-x}{x-8}\right| \leq \dfrac{|x^2-x|}{8-|x|} < \dfrac{|x^2-x|}{7} < \dfrac{|x^2| + |x|}{7} < \dfrac{|x|+|x|}{7} = \dfrac{2|x|}{7} < \dfrac{2}{7}\cdot \dfrac{7\epsilon}{2} = \epsilon$
• How did you think of $\delta=min(1, \frac {7\epsilon}{2})$? – Tim Dec 17 '14 at 20:34
• I want that $|x| < 1 \rightarrow \delta < 1$, and $\dfrac{2|x|}{7} < \epsilon \to |x| < \dfrac{7\epsilon}{2} \to \delta < \dfrac{7\epsilon}{2} \to \delta \leq \text{min}\left(1,\frac{7\epsilon}{2}\right)$. – DeepSea Dec 17 '14 at 20:37