Consider a sequence of functions $f_n:[0,2]\rightarrow\mathbb{R}$ such that $f(0)=0$ and $f(x)=\frac{\sin(x^n)}{x^n}$ for all other $x$. Find $\lim_{n\rightarrow\infty}\int_0^2f_n(x)dx$.
My attempt:
$\sup_{x\in(0,2]}|f_n(x)-f(x)|=\sup_{x\in(0,2]}|\frac{\sin(x^n)}{x^n}-0|$, thus, we have uniform convergence. So, since the interval we are integrating over is finite and $f_n\rightarrow f$ uniformly, then by the Dominated Convergence Theorem, we may pass the limit into the integral. So, we then have $\lim_{n\rightarrow\infty}\int_0^2f_n(x)dx=\int_0^2\lim_{n\rightarrow\infty}f_n(x)dx=\int_0^20dx=0$.
I was wondering if I did this correctly, or if I made a mistake somewhere. In particular, are there any missing small details? I would really appreciate any insight! Thank you.
Best Answer
Your argument is not correct and even the value of the limit is wrong. It is not true that $f_n(x) \to 0$ uniformly on $(0,2]$. In fact $f_n(x) \to 1$ for $0<x<1$ since $x^{n} \to 0$ and $\frac {\sin x} x \to 1$ as $ x \to 0$. For $x>1$ the limit is $0$ and you can apply DCT to show that $\int_1^{2} f_n(x)dx \to 0$. Now $\int_0^{1} f_n(x) dx \to 1$ by DCT since $|\sin t| \leq t$ for all $t \geq 0$. So the right answer is $1$.