# Evaluate limits:$\lim\limits_{n\to\infty}\int_0^nf_n^2\ dm$, $\lim\limits_{n\to\infty}\int_0^nf_n\ dm$ with $f_n(x)=\frac{e^{\sin(x^2/n)}}{1+x}$

So I am working through some practice problems, and on one of them I can't get the second part:

For $x\in(0,\infty)$ and $n\in\{1,2,3,\dots\},$ let $f_n(x)=\frac{e^{\sin\left({x^2/n}\right)}}{1+x}.$ Evaluate with proof: (A) $\lim_{n\rightarrow\infty}\int_0^nf_n^2 \ dm,$ and (B) $\lim_{n\rightarrow\infty}\int_0^nf_n \ dm.$

(A) So I have that $|f_n^2\chi_{[0,n]}|\leq\left|\frac{e^{2\sin\left({x^2/n}\right)}}{(1+x)^2}\right|\chi_{[0,n]}\leq\frac{e^2}{(1+x)^2}.$ We know that $\int_0^\infty \frac{e^2}{(1+x)^2}=\frac{-e^2}{(1+x)}|^\infty_0=e^2<\infty.$ Hence, we have by Dominated Convergence Theorem that $$\lim_{n\rightarrow\infty}\int_0^\infty f_n^2\chi_{[0,n]} dx = \int_0^\infty \lim_{n\rightarrow\infty}f_n^2\chi_{[0,n]}dx = \int_0^\infty\frac{1}{(1+x)^2} dx=1.$$

(B) I can't get the same result because $\int_0^\infty\frac{e}{(1+x)} dx$ does not converge. I am not sure how to approach this part. Any guidance will be appreciated.

• The statement of the problem doesn't seem to match what you're doing in your solution to (A). May 28, 2015 at 7:23
• You mean that I did (A) incorrectly? May 28, 2015 at 7:29
• More likely, you didn't write the question the way you really wanted to. (Note there are no integral signs in it....) May 28, 2015 at 7:30
• Oh, apologizes I will correct that at once. May 28, 2015 at 7:33
• Aha, that's clearer. Hint: $f_n(x) \ge \frac1{e(1+x)}$ for all $n,x$. May 28, 2015 at 7:37

Hint: $\displaystyle f_n(x) \ge \frac1{e(1+x)}$ for all $n,x$.