Evaluate $\lim_{n \to \infty} \int_{0}^{1} x^2 \left(1+\frac{x}{n}\right)^n dx$ Evaluate $$I=\lim_{n \to \infty} \int_{0}^{1} x^2 \left(1+\frac{x}{n}\right)^n dx$$
My try:
we have
$$I=\lim_{ n\to \infty} \frac{1}{n^n} \int_{0}^{1} \left((x+n)^2-2n(x+n)+n^2\right)(x+n)^ndx$$ $\implies$
$$I=\lim_{ n\to \infty} \frac{1}{n^n} \int_{0}^{1}(x+n)^{n+2}-2n(x+n)^{n+1}+n^2(x+n)^n dx$$ $\implies$
$$I=\lim_{ n\to \infty} \frac{1}{n^n} \left(\frac{(n+1)^{n+3}}{n+3}-\frac{2n \times(n+1)^{n+2}}{n+2}+\frac{n^2 \times (n+1)^{n+1}}{n+1}\right)$$ $\implies$
$$I=\lim_{ n\to \infty} (n+1)\frac{(n+1)^{n}}{n^n} \frac{n^2+n+2}{(n+1)(n^2+5n+6)}=e$$
But the answer given is $e-2$
Any thing went wrong?
 A: $$\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x.$$
This convergence is uniform on compact intervals.
Then
$$\lim_{n\to\infty}\int_0^1x^2\left(1+\frac xn\right)^n\,dx
=\int_0^1\lim_{n\to\infty}x^2\left(1+\frac xn\right)^n\,dx
=\int_0^1 x^2e^x\,dx$$
etc. Interchange of limit and integral can be justified by uniform
convergence.
A: Notice that $(1+\frac{x}{n})^n = e^x$. This problem will become a simple integration by parts on $x^2, e^x$ IF we're allowed to change the order of limits and integration. We will justify the swap in order of limit and integration by applying a theorem (The Dominated Convergence Theorem https://en.wikipedia.org/wiki/Dominated_convergence_theorem). 
In particular, let $f_k = x^2(1+\frac{x}{k})^k dx$. Notice that $f_k \rightarrow x^2e^x$ $:=f(x)$ as $k \rightarrow \infty$. Then notice by the taylor series of $e^t$, $e^t = 1 + t + t^2/2! + ....$. Hence if $t > 0$, then $e^t > (1+t)$. So one has $e^{x/n} > (1+\frac{x}{n})$ $\Rightarrow$ $e^{x} > (1+\frac{x}{n})^n$. So we see that $x^2(1+\frac{x}{k})^k \leq x^2e^x$ for $x \in [0,1]$. So we can now apply the dominated convergence theorem one has 
$\lim_{n \rightarrow \infty} \int_{0}^{1} f_n(x) dx = \int_{0}^{1} f(x) dx$ $= \int_{0}^{1} x^2e^x $. Now finish this problem with integration by parts.  
A: I learned a thing or two about the DCT theorem, so let's give it a try. Observe that if $t \in [0,1] \implies \ln(1+t) \le t$ and substitute $t = \dfrac{x}{n}, 0 \le x \le 1, n \ge 1\implies \ln(1+\dfrac{x}{n}) \le \dfrac{x}{n}\implies 1+\dfrac{x}{n} \le e^{\frac{x}{n}}\implies (1+\dfrac{x}{n})^{n} \le e^x\implies |f_n(x)| = x^2(1+\dfrac{x}{n})^n\le x^2e^x=g(x), \text{g is integrable}\implies \displaystyle \lim_{n \to \infty} \displaystyle \int_{0}^1f_n(x)dx =\displaystyle \int_{0}^1\displaystyle \lim_{n\to \infty} f_n(x)=\displaystyle \int_{0}^1 g(x)dx= ...= e-2$ by applying the DCT .
