This is a problem from Problems in Mathematical Analysis: Integration by Kaczor and Nowak:
For a function $f$ continuous on $[0,1]$, find $\lim_{n\to\infty}\int_0^1f(x^n)dx.$
Here is the given solution:
Let $0<\epsilon<1$. Then $\int_0^1f(x^n)dx=\int_0^{1-\epsilon}f(x^n)dx+\int_{1-\epsilon}^1f(x^n)dx$and, by the first mean value theorem, $\int_0^{1-\epsilon}f(x^n)dx=f(\xi^n)(1-\epsilon,\quad\text{where }0\le\xi\le(1-\epsilon).$Thus $\lim_{n\to\infty}f(0)(1-\epsilon).$Moreover, $\left|\int_{1-\epsilon}^1f(x^n)dx\right|\le M\epsilon,\quad\text{where }M=\sup\lbrace|f(x)|:x\in[0,1]\rbrace.$Consequently, $\lim_{n\to\infty}\int_0^1f(x^n)dx=f(0).$
I understand everything except the last part, how do we conclude that $\lim_{n\to\infty}\int_0^1f(x^n)dx=f(0)$?
Anyway, I think we can use the same argument to show that $\lim_{n\to\infty}\int_{1-\epsilon}^1f(x^n)dx=f(0)\epsilon$. If this is correct, why don't we just choose $\epsilon=\frac12$ from the start, to make it simpler?