Proving $\lim_{x \to \infty} \left(1 - \left(1 - \frac{1}{x}\right)^x\right)^x=0$ by using $\lim_{x \to \infty}\left(1-\frac{1}{x}\right)^x=\frac1e$ I'm interested in
$$\lim_{x \to \infty} \left(1 - \left(1 - \frac{1}{x}\right)^x\right)^x \tag1$$
I know this should go to $0$, and the way I want to argue this is that
$$\lim_{x \to \infty} \left(1 - \frac{1}{x}\right)^x = \frac1e \tag2$$
so the entire expression looks like
$$\lim x_{\to \infty} \left(1 - \frac1e\right)^x = 0 \tag3$$
How can I formalize this argument? Or what is another way to find this limit?
 A: For large enough $x$ we have
$$0<1-\left(1 - \frac{1}{x}\right)^x <.9$$
now use sandwich lemma.
A: You could let $x=\frac 1y$ and consider
$$f=\left(1-(1-y)^{\frac{1}{y}}\right)^{\frac{1}{y}}$$ and now compose Taylor series
$$A=(1-y)^{\frac{1}{y}}\implies \log(A)={\frac{1}{y}}\log(1-y)$$
$$\log(A)=-1-\frac{y}{2}-\frac{y^2}{3}+O\left(y^3\right)$$
$$A=e^{\log(A)}=\frac{1}{e}-\frac{y}{2 e}-\frac{5 y^2}{24 e}+O\left(y^3\right)$$
$$B=1-(1-y)^{\frac{1}{y}}=1-A=\left(1-\frac{1}{e}\right)+\frac{y}{2 e}+\frac{5 y^2}{24 e}+O\left(y^3\right)$$
$$f=B^{\frac{1}{y}}\implies \log(f)=\frac 1y \log(B)$$
$$\log(B)=(\log (e-1)-1)+\frac{y}{2 (e-1)}+\frac{(5 e-8) y^2}{24 (e-1)^2}+O\left(y^3\right)$$
$$\log(f)=\frac{\log (e-1)-1}{y}+\frac{1}{2 (e-1)}+\frac{(5 e-8) y}{24 (e-1)^2}+O\left(y^2\right)$$
So, since $\log (e-1)-1 <0$,  $\log(f)\to -\infty$ and $f\to 0$.
A: Here's an example of a solution that uses that fact.
First, we have:
$$\lim_{x \to \infty} \left(1-\left(1-\frac{1}{x}\right)^x \right)^x=$$
Then we can take both the log and the exp of the limit:
$$\lim_{x \to \infty} exp\left(log\left(\left(1-\left(1-\frac{1}{x}\right)^x \right)^x\right)\right)=$$
And using log product rule and passing exp through the limit:
$$exp\left( \lim_{x\to\infty} x\cdot log \left( 1-\left( 1-\frac{1}{x} \right)^x
 \right) \right)=$$
Split the limit and pass log through the limit:
$$exp\left(\left( \lim_{x\to\infty}x \right) \cdot log\left( \lim_{x\to\infty} 1-\left( 1-\frac{1}{x} \right)^x \right) \right)=$$
Now, that we have removed all other factors, the fact you specified can be applied inside $log$!
$$exp\left(\left( \lim_{x\to\infty}x \right) \cdot log\left( 1-\frac{1}{e} \right) \right)=$$
Then, note that $1-\frac{1}{e}<1$ so $log \left( 1-\frac{1}{e} \right) < 0$.
So, as x approaches infinity, the inside of $exp()$ diverges to $-\infty$, so:
$$exp(-\infty) = 0$$
Edit: I did something wrong in (4). I didn't notice that I used the multiplication rule for limits on a divergent limit. Now, instead of splitting the limit, I'm evaluating only part of the limit at a time while leaving another part in x. I'm also not sure if that's allowed. Does this work instead?
$$exp\left( \lim_{x\to\infty} x\cdot log \left( 1-\left( 1-\frac{1}{x} \right)^x \right) \right)=$$
$$exp\left( \lim_{x\to\infty} x\cdot log \left( 1-\frac{1}{e}\right) \right)=$$
$$exp(-\infty) = 0$$
