4
$\begingroup$

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!

$\endgroup$
5
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How did you think of $\delta=min(1, \frac {7\epsilon}{2})$? $\endgroup$ – Tim Dec 17 '14 at 20:34
  • $\begingroup$ I want that $|x| < 1 \rightarrow \delta < 1$, and $\dfrac{2|x|}{7} < \epsilon \to |x| < \dfrac{7\epsilon}{2} \to \delta < \dfrac{7\epsilon}{2} \to \delta \leq \text{min}\left(1,\frac{7\epsilon}{2}\right)$. $\endgroup$ – DeepSea Dec 17 '14 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.