Consider this question:
$$ \lim_{x\to1^-}{{2^{-2}}^{(1-x)}}^{-1} $$
Now, if I directly substitute the value of x, then I get $$ \lim_{x\to1^-}{2^{-2}}^{\infty} $$
All good till now, but here, if $\infty$ is odd then I get the LHS expression, but if $\infty$ is even then I get the RHS expression. And we actually don't know what $\infty$ is.
$$ \lim_{x\to1^-}{2^{-\infty}}\quad\textrm{OR}\quad\lim_{x\to1^-}{2^{\infty}} $$
LHS gives $0$
RHS gives $\infty$.
Hence, the answer according to me is Limit doesn't exist. But according to my book the answer is $0$. What's wrong in my argument?
Note that $\lim_{x\to1^-}(1-x)^{-1}=\infty$. Therefore, $\lim_{x\to1^-}-2^{(1-x)^{-1}}=-\infty$, and so$$\lim_{x\to1^-}2^{-2^{(1-x)^{-1}}}=0.$$