Prove that $\displaystyle\lim_{x \to 2} \dfrac{x+2}{(x-2)^4} = \infty$ using epsilon-delta.

The following proof is provided:

Note that if $x>0$, then $x+2>1$, in which case $\dfrac{x+2}{(x-2)^4} > \dfrac{1}{(x-2)^4}$. If furthermore, $|x-2| < \delta$, then $\dfrac{x+2}{(x-2)^4} > \dfrac{1}{(x-2)^4} > \dfrac{1}{\delta ^4}$. This is larger than a given $N>0$, if $\delta \leq \dfrac{1}{\sqrt[4]{N}}$.

Then it goes on to define delta as $\delta= \min(2,\dfrac{1}{\sqrt[4]{N}}). $ Now I don't understand why the $2$ is there in particular instead of some other number. Usually I understand the necessity of the minimum function, but in this case I cannot grok it for the life of me. Can anyone enlighten me, perhaps by using an example where it goes wrong if there is no $\min(2,..)$?


It is there to guarantee that $x>0$ which is used on the first line.

  • $\begingroup$ I still don't quite get why. How does your choice of $\delta$ have bearing on $x$? $\endgroup$ – YakSal Tafri Feb 21 '17 at 22:50
  • $\begingroup$ It determines the maximum distance $x$ can be away from 2. Since $\delta\leq 2$, $0\leq x\leq 4$. $\endgroup$ – user416426 Feb 21 '17 at 22:54

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.