What is $\lim\limits_{n\rightarrow\infty}\left(1-\left(1-\frac1n\right)^{f(n)}\right)^{2f(n)}$ when $f(n)$ grows faster than $n$? What is $$\lim_{n\rightarrow\infty}\left(1-\left(1-\frac1n\right)^{f(n)}\right)^{2f(n)}$$ when $f(n)$ grows faster than $n$? Is the limit $1$?
How fast should $f(n)$ grow for this to happen?
 A: tl;dr: the limit could be anything in $[0,1]$.
Let $f$ be a function of the form
$$
f(n) = n g(n)
$$
with $g(n)\xrightarrow[n\to\infty]{} \infty$. Then we can write
$$
\left(1-\left(1-\frac1n\right)^{f(n)}\right)^{2f(n)}
= e^{ 2f(n) \ln \left(1-\left(1-\frac1n\right)^{f(n)}\right) }
= e^{ 2n g(n) \ln \left(1-\left(1-\frac1n\right)^{n g(n)}\right) }.
$$
Let's break it down.
$$
\left(1-\frac1n\right)^{n g(n)}
= e^{g(n)\cdot n \ln\left(1-\frac1n\right) }
= e^{g(n) (-1+o(1)) } \xrightarrow[n\to\infty]{} 0
$$
since $g(n)\to\infty$. Therefore, we can use the Taylor expansion of $\ln$ to get:
$$
2n g(n) \ln \left(1-\left(1-\frac1n\right)^{n g(n)}\right)
= - 2n g(n) \left(1-\frac1n\right)^{n g(n)}
= - 2n g(n)e^{-g(n) (1+o(1)) }
$$
This gives us some intuition. This can go, depending on $g(n)$, to either $0$, $-\infty$, or basically any constant $c<0$.
For instance, take


*

*$g(n) = \ln n$: then $- 2n g(n)e^{-g(n) (1+o(1)) } = -2(1+o(1))\ln n\xrightarrow[n\to\infty]{} -\infty$, and the overall limit will be $0$. (After taking the exponential).

*$g(n) = 2\ln n$: then $- 2n g(n)e^{-g(n) (1+o(1)) } = -2(1+o(1))\frac{\ln n}{n}\xrightarrow[n\to\infty]{} 0$, and the overall limit will be $e^0=1$.

*$g(n) = \ln n + \ln \ln n$: then one can check (being a bit more careful in the above, that is in the $o(1)$ term) that $2n g(n) \ln \left(1-\left(1-\frac1n\right)^{n g(n)}\right)\xrightarrow[n\to\infty]{} -2$, and the overall limit will be $e^{-2}$.
A: The first thing that comes to mind is to take a logarithm of the expression you need to find the limit of:
$$
2 f(n) \ln \left[
1 - \left( 1 - {1 \over n} \right)^{f(n)}
\right].
$$
We now want to estimate (the existence of) the limit of 
$$
\left( 1 - {1 \over n} \right)^{f(n)}.
$$
Write
$$
\left( 1 - {1 \over n} \right)^{f(n)} = 
\left\{
\left( 1 - {1 \over n} \right)^{n}
\right\}^{f(n)/n}
$$
and take the logarithm once again:
$$
({f(n)/n}) \ln \left\{
\left( 1 - {1 \over n} \right)^{n}
\right\}
$$
As $n \rightarrow \infty$, the latter logarithm tends to a definite limit (-1), and the ratio $({f(n)/n})$ to infinity (since we are told that $f(n)$ grows faster than $n$.
