Problem with evaluating $\lim_{n \rightarrow \infty} e^{-n-t\sqrt{n}}\cdot \left(e^{e^{\frac{t}{\sqrt{n}}}\cdot n}-1\right)$

Question

As above, I have a problem with evaluating $$\lim_{n \rightarrow \infty} e^{-n-t\sqrt{n}}\cdot \left(e^{e^{\frac{t}{\sqrt{n}}}\cdot n}-1\right).$$ I checked the result in the limit calculator and it should be equal to $e^{\frac{t^2}{2}}$. I have no idea how to obtain it.
My attempt:
$ \lim_{n \rightarrow \infty} e^{-n-t\sqrt{n}}\cdot(e^{e^{\frac{t}{\sqrt{n}}}\cdot n}-1)=\lim_{n \rightarrow \infty} e^{-n-t\sqrt{n}+e^{\frac{t}{\sqrt{n}}}\cdot n}-\lim_{n \rightarrow \infty}e^{-n-t\sqrt{n}}=\lim_{n \rightarrow \infty} e^{-n-t\sqrt{n}+e^{\frac{t}{\sqrt{n}}}\cdot n}-0=?.$
Unfortunately I don't know how to proceed later, because I'm obtaining the limit of the type $0\cdot \infty$ which is of course problematic.
Thanks for help in advance.

Answer 1

Hint use Taylor expansion of $\displaystyle e^{\frac{t}{\sqrt{n}}}$.

$$\exp\left(\frac{t}{\sqrt{n}}\right) \approx 1 + \frac{t}{\sqrt{n}} + \frac{t^2}{2n} + O\left(\frac{t^3}{n^{3/2}}\right)$$

Answer 2

As an alternative, by $\log (1+x)=\frac{t}{\sqrt{n}} \implies x=e^{\frac{t}{\sqrt{n}}}-1 \to 0$ and $n=\frac{t^2}{\log^2(1+x)}$ we obtain

$$e^{-n-t\sqrt{n}}\cdot \left(e^{e^{\frac{t}{\sqrt{n}}}\cdot n}-1\right)=e^{-\frac{t^2}{\log^2(1+x)}-\frac{t^2}{\log(1+x)}}\cdot \left(e^{\frac{t^2(1+x)}{\log^2(1+x)}}-1\right)=$$

$$=e^{\frac{xt^2}{\log^2(1+x)}-\frac{t^2}{\log(1+x)}}-e^{-\frac{t^2}{\log^2(1+x)}-\frac{t^2}{\log(1+x)}} \to e^{\frac{t^2}2}-0= e^{\frac{t^2}2} $$

indeed

$$\frac{xt^2}{\log^2(1+x)}-\frac{t^2}{\log(1+x)}=t^2\frac{x-\log(1+x)}{\log^2(1+x)}=t^2\frac{x-\log(1+x)}{x^2}\frac{x^2}{\log^2(1+x)}\to \frac{t^2}2$$

since by standard limits

$$\frac{x}{\log(1+x)}\to 1$$

and by this result

$$\frac{x-\log(1+x)}{x^2}\to \frac12$$