$\epsilon$-$\delta$ proof that $\lim\limits_{x \to 1} 2^{-\frac{1}{(x-1)^2}} = 0$

Question

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 $\delta$.

Answer 1

Taking your work:

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

$$|x-1|<\frac1{\sqrt{-\log_2\epsilon}}$$

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

Answer 2

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 $\delta>0$.