(a) Find the pointwise limit function $f$ on $[−1, 1]$.
(b) Show that $\lim\limits_{n \to \infty} \int_{-1}^1 f_n(x)dx$ exists. Is it equal to $\int_{-1}^1 f(x)dx$?
This is my solution: For part a, I got that the limit of the sequence of the function, is the piecewise function: (Edit, my answer for part a was wrong so I have updated):
$f(x) = \begin{cases} {1/x}, & \text{if $x$ does not equal to 0} \\ 0, & \text{if $x = 0$} \end{cases}$
I am stuck on part b. I am not sure how to show that $\lim\limits_{n \to \infty} \int_{-1}^1 f_n(x)dx$ exists. Do I show that the derivative of the $f_n(x)$ exists and its continuous of $[-1,1]$?
As for the second part of the question "is it equal to $\int_{-1}^1 f(x)dx$?". By the integration and uniform limit theorem, since $f_n(x)$ does not converge uniformly to $f$ on [a,b], then $\lim\limits_{n \to \infty} \int_{-1}^1 f_n(x)dx$ does not equal to $\int_{-1}^1 f(x)dx$? Is the reasoning right?