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!

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}
