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

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 $$e.$$

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 $\lim\limits_{n \to \infty} \int\limits_0^1 n x^n g(x) dx = g(1)$

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.