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?
 A: Alternatively we could compute this limit directly via the substitution $u = x^{n+1}$
$$\lim_{n\to\infty} \frac{n}{n+1}\int_0^1e^{u^{\frac{2}{n+1}}}\:du \to \int_0^1e\:du = e$$
by dominated convergence.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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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.
A: A bit late answer but maybe worth noting it.
Partial integration gives
$$I_n :=\int_0^1 \underbrace{nx^{n-1}}_{u'}\cdot\underbrace{xe^{x^2}}_{v}dx= \left.x^{n+1}e^{x^2}\right|_0^1- \underbrace{\int_0^1 x^n(1+2x^2)e^{x^2}dx}_{J_n=}=e-J_n$$
Now, $J_n$ can be easily estimated as follows
$$0\leq J_n \leq 3e\int_0^1x^ndx=\frac{3e}{n+1}\stackrel{n\to \infty}{\longrightarrow}0$$
Hence, $I_n \stackrel{n\to \infty}{\longrightarrow} e$.
A: As the Taylor series of $e^{x^2}$ converges uniformly in $[0,1]$,
$$\int_0^1n x^n\sum_{k=0}^\infty\frac{x^{2k}}{k!}dx=\int_0^1 n\sum_{k=0}^\infty \frac{x^{2k+n}}{k!}dx=\sum_{k=0}^\infty\frac n{(2k+n+1)k!}\\=\sum_{k=0}^\infty\frac 1{k!}-\sum_{k=0}^\infty\frac{2k+1}{(2k+n+1)k!}.$$
The second term vanishes because
$$\sum_{k=0}^\infty\frac{2k+1}{(2k+n+1)k!}<\frac 1n\sum_{k=0}^\infty\frac{2k+1}{k!}.$$

 Hence $$e.$$

A: Your proof is incorrect. The issue is that the $c$ which you choose may depend on $n$.
It turns out that the correct answer is in fact $e$. This is because for any continuous $f : [0, 1] \to \mathbb{R}$, we have
$\lim\limits_{n \to \infty} \int\limits_0^1 n x^n f(x) dx = f(1)$
This follows from the Stone-Weierstrass theorem as follows:
First, we prove that $\lim\limits_{n \to \infty} \int\limits_0^1 n x^n f(x) dx = f(1)$ for every $f$ of the form $f(x) = x^m$. We then extend this to all $f$ polynomial quite easily.
Suppose now that we have continuous $g : [0, 1] \to \mathbb{R}$. Given arbitrary $\epsilon > 0$, let $w = \frac{\epsilon}{3}$. Take polynomial $f$ s.t. $|f - g| < w$ (uniform norm) which is possible by Stone-Weierstrass, and take $N$ s.t. for all $n \geq N$, $\left|\int\limits_0^1 n x^n f(x) dx - f(1)\right| < w$. Then we have
\begin{equation}
\begin{split}
\left| \int\limits_0^1 n x^n g(x) dx - g(1) \right| 
&\leq \left|\int\limits_0^1 n x^n g(x) dx - \int\limits_0^1 n x^n f(x) dx\right| + \left|\int\limits_0^1 n x^n f(x) dx - f(1)\right| + \left|g(1) - f(1)\right| \\[10pt]
&= \left|\int\limits_0^1 n x^n (g(x) - f(x)) dx \right| + |g(1) - f(1)| + \left|\int\limits_0^1 n x^n f(x) dx - f(1)\right| \\[10pt]
&< \left|\int\limits_0^1 n x^n (g(x) - f(x)) dx \right| + w + w \\[6pt]
&\leq \int\limits_0^1 n x^n \left|g(x) - f(x)\right| dx + 2w \\
&\leq w \int\limits_0^1 n x^n dx + 2w \\
&= w \frac{n}{n + 1} + 2w \\[6pt]
&<3w \\[6pt]
&= \epsilon
\end{split}
\end{equation}
And therefore $\lim\limits_{n \to \infty} \int\limits_0^1 n x^n g(x) dx = g(1)$
