I got stuck at limit problem 
$$\lim_{n\to\infty} n\bigg[e^{\frac x{\sqrt n}}-\frac x{\sqrt n}-1\bigg] = \frac{x^2}{2}$$

I'm not sure how to solve it.
hope somebody could help me!
_ Is there any way to see solutions for limit problem?
 A: By the change of variable $t:=x/\sqrt n$ you can write
$$
\lim_{n\to\infty}n\left(e^{x/\sqrt n}-\frac x{\sqrt n}-1\right)=
x^2\lim_{t\to0^+}\frac{e^t-t-1}{t^2}$$ and the dependency on $x$ is gone.
Now by L'Hospital, twice,
$$\lim_{t\to0^+}\frac{e^t-t-1}{t^2}=\lim_{t\to0^+}\frac{e^t-1}{2t}=\lim_{t\to0^+}\frac{e^t}2.$$

Without L'Hospital, assuming the limit exists,
$$2L-L=2\lim_{2t\to0^+}\frac{e^{2t}-2t-1}{4t^2}-\lim_{t\to0^+}\frac{e^t-t-1}{t^2}=\lim_{t\to0^+}\frac{2e^{2t}-4e^t+2}{4t^2}=\frac12\lim_{t\to0^+}\left(\frac{e^t-1}t\right)^2$$ and the last limit is known to have the value $1^2$.
A: Sice $e^{z/\sqrt{n}}$ is analytic we can consider
$$
e^{z/\sqrt{n}} = 1 + \frac{z}{\sqrt{n}} + \sum_{m=2}^\infty \frac{z^m}{\sqrt{n^m}m!}
$$
So
$$
n\left(e^{z/\sqrt{n}} - 1 - \frac{z}{\sqrt{n}}\right) = \sum_{m=2}^\infty \frac{z^n}{n^{m/2-1}m!} = \frac{z^2}{2} + \sum_{m=3}^\infty \frac{z^m}{n^{m/2-1}m!}
$$
Taking limit and changing limit and sum (you can check easily dominate convergence) have the result.
A: For $x\neq 0$
$$\lim_{n\to\infty} n\bigg[e^{\frac x{\sqrt n}}-\frac x{\sqrt n}-1\bigg] \stackrel{y=\frac{x}{\sqrt{n}}}{=} x^2 \lim_{y\to0} \frac{e^y-y-1}{y^2} = x^2 \lim_{y\to0} \frac{e^y-1}{2y} = x^2 \lim_{y\to0} \frac{e^y}{2} = \frac{x^2}{2}$$
