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?

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 $1^2$.
