Result of $\lim_{n\to\infty} (1+ \sqrt{n+1} -\sqrt{n})^{\sqrt{n}}$

Question

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.

Answer 1

We have that

$$(1+ \sqrt{n+1} -\sqrt{n})^{\sqrt{n}} =\left(1+ (\sqrt{n+1} -\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\right)^{\sqrt{n}} =\left(1+ \frac{1}{\sqrt{n+1}+\sqrt{n}}\right)^{\sqrt{n}} =\left[\left(1+ \frac{1}{\sqrt{n+1}+\sqrt{n}}\right)^{\sqrt{n+1}+\sqrt{n}}\right]^{\frac{\sqrt n}{\sqrt{n+1}+\sqrt{n}}}\to e^\frac12$$

Answer 2

Hint: Calculate $$ (\sqrt{n+1}-\sqrt{n})\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} $$ Then note that $$ 2\sqrt n\leq \sqrt{n+1}+\sqrt n\leq2\sqrt{n+1} $$

Answer 3

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}