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(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?

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    $\begingroup$ Unless there is a typo the pointwise limit is $\frac 1 x$ for $x \neq 0$ and $\int f(x)dx$ does not even exist. $\endgroup$ – Kavi Rama Murthy May 20 at 8:10
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    $\begingroup$ See math.stackexchange.com/questions/1230653/… for the limit. $\endgroup$ – Arnaud D. May 20 at 8:18
  • $\begingroup$ @KaviRamaMurthy Yes, my bad I did it the other way round so I updated my answer for part a. But I am struggling to show the first half of part b, the integral of the $f_n(x)$ $\endgroup$ – codelearner May 20 at 8:31
  • $\begingroup$ @ArnaudD. yes, Thank you. I realised the mistake I made for part a. $\endgroup$ – codelearner May 20 at 8:31
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    $\begingroup$ For the integral, note that each function $f_n$ is odd. $\endgroup$ – Arnaud D. May 20 at 8:34

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