I have to use lebesgue dominated convergence theorem to prove that
$ \lim_{n\rightarrow\infty}\int_0^\infty \left[1+ \frac{\ln(x + n^2)}{n^{1/2}}\sin(x^2) + \cos\left(\frac{1}{n+x}\right)\right] e^{-x/5} \, d\lambda\ = 5 $
Do I have to find the function $g$ which bounds all $f_n$? [since it already says to prove using the dominated convergence theorem]
What I have done so far is:
$\lim_{n\rightarrow\infty}\int_0^\infty \left[1+ \frac{\ln(x + n^2)}{n^{1/2}}\sin(x^2) + \cos\left(\frac{1}{n+x}\right)\right] e^{-x/5} \, d\lambda $ $ = \int_0^\infty \lim_{n\rightarrow\infty}\left[1+ \frac{\ln(x + n^2)}{n^{1/2}}\sin(x^2) + \cos\left(\frac{1}{n+x}\right)\right] e^{-x/5} \, d\lambda$
Then,
$\frac{\ln(x+n^2)}{n^{1/2}}=\frac{\ln\left(n^2\left(1+\frac{x}{n^2}\right)\right)}{n^{1/2}} = \frac{2\ln n +\ln\left(1+\frac{x}{n^2}\right)}{n^{1/2}}$
I used the maclaurin expansion of $\ln(1+X)$ to prove that $\lim_{n\rightarrow\infty}\frac{\ln(1+\frac{x}{n^2})}{n^{1/2}}=0$
I compared $\lim_{n\rightarrow\infty}\frac{\ln{n}}{n^{1/2}}$ to $\lim_{x\rightarrow\infty}\frac{\ln{x}}{x^{1/2}}$ and used l'Hopital's rule to prove that the limit is zero since I found such an example on youtube.
I don't know whether it is good. Also, when I use maclaurin's expansion of $\cos(X)$ and take $\lim_{n\rightarrow\infty}\cos(\frac{1}{n+x})$, I get it to be $1$.
Then I get $10$ as final answer.