Evaluating the limit $1^\infty$ I am trying to evaluate this limit: 

$$\lim_{n\to\infty}\left[e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^n\right].$$ 

I saw it is a limit of $1^\infty$ type and I tried to evaluate it like this: $(1-\frac{1}{\sqrt{n}})^n=e^\frac{-n}{\sqrt{n}}=e^{-\sqrt{n}}$ and that means the limit equals $e^0=1$. But the answer in my textbook is $\frac{1}{\sqrt{e}}$. Am I not allowed to split the limit like that?
 A: It's  simpler to find he limit of the log first, using Taylor's formula at order $2$:
\begin{align}
\log\Biggl(e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^n\Biggr)&=\sqrt n+n\log\biggl(1-\frac1{\sqrt n}\biggr)=\sqrt n-n\biggl(\frac1{\sqrt n}+\frac1{2n}+o\Bigl(\frac1{ n}\Bigr)\biggr)\\&=-\frac12+o(1).
\end{align}
So the log tends to $-\frac12$ when $n$ tends to $\infty$, and we conclude that
$$\lim_{n\to\infty}\log\Biggl(e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^n\Biggr)=\frac1{\sqrt{\mathrm e}}.$$
A: $$\lim_{n\rightarrow+\infty}e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^n=\lim_{n\rightarrow+\infty}e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}\cdot\frac{n}{-\sqrt{n}}}=$$
$$=\lim_{n\rightarrow+\infty}\left(\frac{e}{\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}}}\right)^{\sqrt{n}}=$$
$$=\lim_{n\rightarrow+\infty}\left(1+\frac{e}{\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}}}-1\right)^{\frac{1}{\left(\frac{e}{\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}}}-1\right)}\cdot\sqrt{n}\left(\frac{e}{\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}}}-1\right)}=$$
$$=e^{\lim\limits_{n\rightarrow+\infty}\sqrt{n}\left(\frac{e}{\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}}}-1\right)}=e^{-\frac{1}{e}\lim\limits_{x\rightarrow0}\frac{e-(1+x)^{\frac{1}{x}}}{x}}$$ and the rest it is the L'Hôpital's rule.
A: We have $n\ln(1-\frac{1}{\sqrt{n}})-\sqrt{n}=-\frac{1}{2}+\mathcal{O}(\frac{1}{\sqrt{n}})$, so the limit is $e^{-1/2}$ as required.
