# Let $f_n(x) = {nx\over 1+nx^2}$ on the domain $[-1,1]$

(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?

• 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. – Kavi Rama Murthy May 20 at 8:10
• See math.stackexchange.com/questions/1230653/… for the limit. – Arnaud D. May 20 at 8:18
• @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 $f_n(x)$ – codelearner May 20 at 8:31
• @ArnaudD. yes, Thank you. I realised the mistake I made for part a. – codelearner May 20 at 8:31
• For the integral, note that each function $f_n$ is odd. – Arnaud D. May 20 at 8:34