Solve the following limit $\lim _{n\to \infty }\left( e^{\sqrt{n+1} -\sqrt{n}} -1\right) \times \sqrt{n}$ can someone try to solve this limit? $$\lim _{n\to \infty }\left( e^{\sqrt{n+1} -\sqrt{n}} -1\right) \times \sqrt{n}$$
I tried to solve it like $\lim _{n\to \infty }\left( e^{\sqrt{n+1} -\sqrt{n}} -1\right) \times \lim _{n\to \infty }\sqrt{n}$ but it always tends to 0, but in wolfram the result is $\frac{1}{2}$ so Im a bit confused, thanks!
 A: Call $x=1/\sqrt{n}$, so your limit becomes a limit as $x\to 0$. It becomes
$$
\lim_{x\to 0}\frac{e^{\frac{(1+x^2)^{1/2}-1}{x}}-1}{x}
$$
Expanding as a power series:
$$
=\lim_{x\to 0}\frac{e^{\frac{x}{2}+\dots}-1}{x}=\lim_{x\to 0}\frac{\frac{x}{2}+\dots}{x}=\frac 1{2}.
$$
This is way faster than using L'Hopital's rule. No derivatives are needed, just elementary expansions.
A: Without using L'Hospital and knowing
$$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0, n\to\infty$$
we can write, take in account my comment,
$$\lim\limits _{n\to \infty }\left( e^{\sqrt{n+1} -\sqrt{n}} -1\right) \times \sqrt{n} =\lim\limits _{n\to \infty } \frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{2}$$
A: Note that $$ \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} +\sqrt{n}} $$
Let us denote this expression $x_n$. We have $\lim_{n\to\infty} x_n = 0$. Knowing that $\lim_{x\to0} \frac{e^x-1}{x} =1$, we get
\begin{align} & \lim_{n\to\infty} (e^{\sqrt{n+1} - \sqrt{n}} - 1)\sqrt{n} = \\
&=\lim_{n\to\infty} \frac{e^{x_n} - 1}{x_n} x_n \sqrt{n} = \\
&=\lim_{n\to\infty} \frac{e^{x_n} - 1}{x_n} \frac{\sqrt{n}}{\sqrt{n+1} +\sqrt{n}} = \\
&=\lim_{n\to\infty} \frac{e^{x_n} - 1}{x_n} \cdot \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \\
&=\lim_{x\to 0} \frac{e^{x} - 1}{x} \cdot \lim_{n\to\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \\
&= 1 \cdot \frac12 = \frac12 \end{align}
A: When you separate it like that, you get something that behaves like "$0 \times \infty$", which is indeterminate. So instead, write
$$\lim_{n \to \infty} (e^{\sqrt{n+1}-\sqrt{n}} - 1)\sqrt{n} = \lim_{n \to \infty} \frac{(e^{\sqrt{n+1}-\sqrt{n}} - 1)}{1/\sqrt{n}}$$
Notice that the numerator goes to $0$ because $\sqrt{n+1}-\sqrt{n} \to 0$ as $n\to \infty$, and the denominator goes to $0$ as well. Use L-Hospital.
The derivatives start out looking messy, but you will see a lot of simplifications as you let $n\to \infty$ (realize $\sqrt{n+1} \sim \sqrt{n}$ as $n$ grows larger).
