Finding limit of $\frac{2}{2^{\frac{1}{2^n}}}$ I'm trying to find the limit of $\frac{2}{2^{\frac{1}{2^n}}}$ by finding 2 limits smaller and greater than the denominator and thus find the limit of the denominator and thus find the limit of the entire thing. can't find the 2 sequences needed for that. I need to use squeeze theorem to find the limit of the denominator.
 A: We have that by continuity
$$\frac{1}{2^n} \to 0 \implies\frac{2}{2^{\frac{1}{2^n}}} \to \frac{2}{2^{0}}=2$$
As an alternative by squeeze theorem we have that
$$0\le\frac{1}{2^n}\le \frac1n$$
and therefore
$$\frac{2}{2^{0}}\le \frac{2}{2^{\frac{1}{2^n}}}\le \frac{2}{2^{\frac{1}{n}}}$$
and we can conclude that the limit is one sinceboth RHS and LHS tends to $2$ but in this case we are using the same result by continuity therefore it seems a not effective method in this case.
A: As $n$ becomes large $\frac{1}{2^n}$ becomes very small and positive. So $2^{\frac{1}{2^n}}$ becomes very close to $1$ as $n$ becomes large. So overall $\frac{2}{2^{\frac{1}{2^n}}}$ approaches $\frac{2}{1} = 2$ as $n$ becomes large.
A: A bit confused as to why your professor insists on using squeeze theorem, but I guess one way this could work for the denominator is $$2^{\frac{1}{n!}}\leq 2^{\frac{1}{2^{n}}}\leq2^{\frac{1}{n}}$$
Which, if you want to see in action check out this Desmos link: https://www.desmos.com/calculator/ngfb9ds7dc
Anyways, if we take limits of all of these we get $$\lim_{n\rightarrow \infty} 2^{\frac{1}{n!}}\leq\lim_{n\rightarrow \infty} 2^{\frac{1}{2^{n}}}\leq\lim_{n\rightarrow \infty}2^{\frac{1}{n}}$$
And then 
$$1\leq\lim_{n\rightarrow \infty} 2^{\frac{1}{2^{n}}}\leq1$$
so $$lim_{n\rightarrow \infty} 2^{\frac{1}{2^{n}}} = 1$$ By squeeze theorem
