I am trying to show that $\lim\limits_{x \to 1} 2^{-\frac{1}{(x-1)^2}} = 0$ using the $\epsilon$-$\delta$ definition. So far I have done the following: \begin{align*} && \big|2^{-\frac{1}{(x-1)^2}} - 0\big| & < \epsilon\\ \iff && 2^{-\frac{1}{(x-1)^2}} & < \epsilon\\ \iff && - \frac{1}{(x-1)^2} & < \log_2\epsilon, \end{align*} but I can't really see how I can end up with $|x-1| < \delta$ for some $\delta$.

  • $\begingroup$ Why can't you multiply over your positive denominator and take roots? $\endgroup$ – JuliusL33t Apr 21 '18 at 15:52
  • $\begingroup$ @JuliusL33t Won't that give me $\sqrt{\frac{-1}{\log_2\epsilon}} < |x-1|$, i.e. a lower-bound for $|x-1|$ instead of an upper-bound? $\endgroup$ – Luke Collins Apr 21 '18 at 15:54
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    $\begingroup$ suppose $0<\epsilon<1$, then $\log_2 \epsilon<0$. So \begin{align*} && \big|2^{-\frac{1}{(x-1)^2}} - 0\big| & < \epsilon\\ \iff && 2^{-\frac{1}{(x-1)^2}} & < \epsilon\\ \iff && - \frac{1}{(x-1)^2} & < \log_2\epsilon \iff && |x-1| &< \sqrt{\frac{-1}{\log_2\epsilon}} \end{align*} $\endgroup$ – Riemann Apr 21 '18 at 16:02

Taking your work:

$$-\frac1{(x-1)^2}<\log_2\epsilon\implies\frac1{(x-1)^2}>-\log_2\implies (x-1)^2<-\frac1{\log_2\epsilon}\implies$$


Observe that $\;\log_2\epsilon<0\;$ for small $\;\epsilon>0\;$ . Take it from here.


Note that when $\epsilon\in(0,1)$ setting $K:=-\log_2\epsilon$ we have that $K\in (0,\infty)$ so your inequality is equivalent to show that for any chosen $K>0$ there is a $\delta >0$ such that

$$\frac1{(x-1)^2}>K,\quad\text{when }0<|x-1|<\delta$$

What if you choose $\delta=\min\{K^{-1},1/2\}$?

P.S.: for $\epsilon\ge 1$ you can just choose any $\delta>0$.


From here assuming $\epsilon<1\implies \log_2\epsilon$

$$- \frac{1}{(x-1)^2} < \log_2\epsilon \iff (x-1)^2<- \frac{1}{\log_2\epsilon}\iff (x-1)^2< \frac{1}{\log_2 \frac1{\epsilon}}\iff |x-1|<\sqrt{ \frac{1}{\log_2 \frac1{\epsilon}}}$$


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