Result of $\lim_{n\to\infty} (1+ \sqrt{n+1} -\sqrt{n})^{\sqrt{n}}$ The task is to $\lim_{n\to\infty} (1+ \sqrt{n+1} - \sqrt{n})^{\sqrt{n}}$. From looking at it it will probably by $e$ type of limit, but I don't know how to continue.
Thank you for your help.
 A: Hint: Calculate
$$
(\sqrt{n+1}-\sqrt{n})\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}
$$
Then note that
$$
2\sqrt n\leq \sqrt{n+1}+\sqrt n\leq2\sqrt{n+1}
$$
A: We have that
$$(1+ \sqrt{n+1} -\sqrt{n})^{\sqrt{n}}
=\left(1+ (\sqrt{n+1} -\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\right)^{\sqrt{n}}
=\left(1+ \frac{1}{\sqrt{n+1}+\sqrt{n}}\right)^{\sqrt{n}}
=\left[\left(1+ \frac{1}{\sqrt{n+1}+\sqrt{n}}\right)^{\sqrt{n+1}+\sqrt{n}}\right]^{\frac{\sqrt n}{\sqrt{n+1}+\sqrt{n}}}\to e^\frac12$$
A: Hint:
Use some asymptotic calculus to compute the limit of the log:
\begin{align}
\sqrt{n+1}-\sqrt n&=\sqrt n\biggl(\sqrt{1+\frac 1n}-1\biggr)=\sqrt n\biggl(\not1+\frac 1{2n}+o\Bigl(\frac1n\Bigr)-\not1\biggr) \\
&= \frac 1{2\sqrt n}+o\Bigl(\frac1{\sqrt n}\Bigr)\\
\text{so}\quad\sqrt n\,\log\bigl(1+\sqrt{n+1}-\sqrt n\bigr)&=\sqrt n\,\log\biggl(1+\frac 1{2\sqrt n}+o\Bigl(\frac1{\sqrt n}\Bigr)\biggr)\\
&=\sqrt n\,\biggl(\frac 1{2\sqrt n}+o\Bigl(\frac1{\sqrt n}\Bigr)\biggr)\\
&=\dots.
\end{align}
