<s>[INST] Limit of $(1-\frac{1}{2^n})^n$. What is the limit of the sequence $(1-\frac{1}{2^n})^n$?
Let $a_n:=(1-\frac{1}{2^n})^n$ then $\ \ln \ a_n=n\ \ln(1-\frac{1}{2^n})$
$\implies \lim \limits_{n\rightarrow \infty} \ln\ a_n=\lim\limits_{n \rightarrow \infty} n\ \ln(1-\frac{1}{2^n})$
$\implies \ln \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \frac{\ln(1-\frac{1}{2^n})}{\frac{1}{n}}$
$\implies \ln \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \frac{\frac{ln2}{2^n(1-2^n)}}{\frac{-1}{n^2}}  [\frac{0}{0} \ \ \text{form}]$
I am stuck here, what should I do after this?
 [/INST] Continuing where you left:
$$  \frac{ \frac{\ln2}{2^n (1 -2^n)}  }{\frac{-1}{n^2}} = - \frac{n^2 \ln2}{2^n(1-2^n)} = \frac{n^2 \ln2}{2^n(2^n - 1)}$$
It's pretty obvious / not hard to prove that this tends to $0$ as $n$ tends to infinity ("exponent beats polynomial") and so your limit should go to $e^0 = 1$
Another approach is to "fix" the exponent so it would suit the limit of $e$:
$$ \lim_{n \rightarrow \infty}( 1 - \frac{1}{2^n})^n = \lim_{n \rightarrow \infty}( \overbrace{(1 + \frac{-1}{2^n})^{2^n}}^{\text{1/e}})^{ \overbrace{\frac{n}{2^n}}^{\text{Tends to 0}}}$$
</s>