Find the limit $\lim_{n\to\infty} \bigl(1+\frac{1}{n}\bigr)^{n^2}\cdot\frac{1}{e^n}$ 
Find $\displaystyle\lim_{n\to\infty} \Bigl(1+\dfrac{1}{n}\Bigr)^{n^2}\cdot\dfrac{1}{e^n}$

We have:
$$
\lim_{n\to\infty} \Bigl(1+\dfrac{1}{n}\Bigr)^{n^2}\cdot\dfrac{1}{e^n}=
\lim_{n\to\infty}\dfrac{e^{n^2\ln\bigl(1+\frac{1}{n}\bigr)}}{e^n}=
\lim_{n\to\infty} e^{n^2\ln\bigl(1+\frac{1}{n}\bigr)-n}
$$
Then I have no idea. Can anyone help me please?

The idea of reducing everything to a single exponential is good. Now you have to evaluate
$$
\lim_{n\to\infty}\Bigl(n^2\ln\Bigl(1+\frac{1}{n}\Bigr)-n\Bigr)
$$
You can try computing instead (with $1/x=t$ after the equality)
$$
\lim_{x\to\infty}\Bigl(x^2\ln\Bigl(1+\frac{1}{x}\Bigr)-x\Bigr)
=
\lim_{t\to0^+}\frac{\ln(1+t)-t}{t^2}=
\lim_{t\to0^+}\frac{t-t^2/2+o(t^2)-t}{t^2}=-\frac{1}{2}
$$
Instead of a Taylor expansion you can use l'Hôpital:
$$
\lim_{t\to0^+}\frac{\ln(1+t)-t}{t^2}=
\lim_{t\to0^+}\frac{\dfrac{1}{1+t}-1}{2t}
$$
