TITLE: Integral $\lim_{n \rightarrow \infty} \int_{0}^{1} \sqrt{\frac{1}{x}+n^2x^{2n}}dx$
QUESTION [7 upvotes]: Evaluate $\lim_{n \rightarrow \infty} \int_{0}^{1} \sqrt{\frac{1}{x}+n^2x^{2n}}dx$.
The integral is non asymptomatic. The convergence is non uniform at both $x=0$ and $x=1$. I'm not sure how to proceed.

REPLY [7 votes]: We have
\begin{align}
I_n &= \int_0^1 \sqrt{\frac{1}{x} + n^2 x^{2n}} \mathrm{d} x \\
&\le \int_0^1 \left(\sqrt{\frac{1}{x}} + nx^n\right) \mathrm{d} x \\
&\le \int_0^1 \left(\sqrt{\frac{1}{x}} + (n+1)x^n\right) \mathrm{d} x \\
&= \left(2\sqrt{x} + x^{n+1}\right)\Big\vert_0^1 \\
&= 3.
\end{align}
Also, we have
\begin{align}
I_n &= \int_0^{1-\frac{1}{\sqrt{n}}} \sqrt{\frac{1}{x} + n^2 x^{2n}}\ \mathrm{d} x 
+ \int_{1-\frac{1}{\sqrt{n}}}^1 \sqrt{\frac{1}{x} + n^2 x^{2n}}\ \mathrm{d} x\\
&\ge \int_0^{1-\frac{1}{\sqrt{n}}} \sqrt{\frac{1}{x}}\ \mathrm{d} x
+ \int_{1-\frac{1}{\sqrt{n}}}^1 nx^n \mathrm{d} x \\
&= 2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right].
\end{align}
Combining the above results, we have
$$2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right] \le I_n \le 3.$$
Note that
$$\lim_{n\to \infty} 2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right] = 3.$$
By the squeeze theorem, we have
$\lim_{n\to \infty} I_n = 3$. We are done.