I was Stumped by this Challenging Integration/Limit Problem Can someone show me how I do the following:
\begin{align}
I_n & = \int_0^1 \sqrt{ \frac 1 x + n^2 x^{2n} } \ \mathrm dx \\[8pt] & \lim_{n\rightarrow \infty } I_n = \text{ ?}
\end{align}
 A: Split the interval of integral into two subintervals
$$[0,1]=[0,1-\epsilon]+[1-\epsilon,1] \ \ \ \ \ \ \ \epsilon \rightarrow 0^{+}$$
For very large $n$ in the first interval we have $\frac{1}{x} \gg n^2 x^{2n}$ and in the second $n^2 x^{2n} \gg \frac{1}{x}$.
The integral becomes
$$I=\int_{0}^{1-\epsilon} \frac{1}{\sqrt{x}} \text{d} x+\int_{1-\epsilon}^{1} n x^n \text{d}x $$
$$I=2\sqrt{1-\epsilon}+\frac{n}{n+1} (1-(1-\epsilon)^{n+1})$$
Taking the limit of $I$ with $\epsilon \rightarrow 0^{+}$ and $n \rightarrow \infty $ gives $I=3$.
A: For any $b \in (0,1),$ we have
$$\int_0^b \frac{dx}{\sqrt x}+ \int_b^1 nx^n\,dx \le I_n.$$
Let $n\to \infty$ to see
$$\int_0^b \frac{dx}{\sqrt x}+ 1 \le \liminf I_n.$$
Now let $b\to 1$ to see
$$\int_0^1 \frac{dx}{\sqrt x}+ 1 = 3 \le \liminf I_n.$$
For an estimate from above, note that for $u,v\ge 0, \sqrt {u+v} \le \sqrt u + \sqrt v.$ Thus
$$I_n \le \int_0^1 (\frac{1}{\sqrt x}+ nx^n)\,dx.$$
Let $n\to \infty$ to see
$$\limsup I_n \le \int_0^1 \frac{dx}{\sqrt x} + 1 = 3.$$
This shows $\lim I_n = 3.$
