<s>[INST] 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!   
 [/INST] For any $\epsilon > 0$, choose $\delta = \text{min}\left(1,\frac{7\epsilon}{2}\right)$, then: if $0 < |x| < \delta$ then $\left|\dfrac{x^2-8}{x-8} - 1\right| = \left|\dfrac{x^2-x}{x-8}\right| \leq \dfrac{|x^2-x|}{8-|x|} < \dfrac{|x^2-x|}{7} < \dfrac{|x^2| + |x|}{7} < \dfrac{|x|+|x|}{7} = \dfrac{2|x|}{7} < \dfrac{2}{7}\cdot \dfrac{7\epsilon}{2} = \epsilon$
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