Evaluate $\lim _{n\to \infty }\int _{0}^{1}nx^ne^{x^2}dx$ 
Evaluate $\lim _{n\to \infty }\int _{0}^{1}nx^ne^{x^2}dx.$

I applied the mean value thorem of integral to $\int _{0}^{1}nx^ne^{x^2}dx.$ We get $c\in (0,1):$
$$\int _{0}^{1}nx^ne^{x^2}dx=(1-0)nc^ne^{c^2}.$$ Taking limit ($\lim_{n\to \infty}$)on the both side,
We get, $$\lim_{n\to \infty}\int _{0}^{1}nx^ne^{x^2}dx=\lim_{n\to \infty} nc^ne^{c^2}=0.$$
My answer in the examination was wrong. I don't know the correct answer. Where is my mistake?

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\begin{align}
\lim _{n \to \infty}\int_{0}^{1}nx^{n}\expo{x^{2}}\dd x & =
\lim _{n \to \infty}\bracks{n\int_{0}^{1}
\exp\pars{n\ln\pars{1 - x} + \pars{1 - x}^{2}}\dd x}
\\[5mm] & =
\lim _{n \to \infty}\bracks{n\int_{0}^{\infty}
\expo{1 -\pars{n + 2}x}\dd x}
\\[5mm] & =
\expo{}\lim _{n \to \infty}{n \over n + 2} =
\bbx{\large\expo{}} \\ &
\end{align}
See Laplace's Method.
