Evaluation of \( \displaystyle \lim_{n\rightarrow \infty}\bigg[\bigg(\ln(n^2+n^{-2})\bigg)^{-1}\cdot \int^{\pi n}_{1}\frac{\sin^2(5x)}{x}dx\bigg] \)
Note: I'm assuming here that \( n\in \mathbb{N} \), that is, that this refers to the limit of a sequence.
I found a way, maybe not the simpler nor faster but it works. Let \( b_n:=\ln(n^2+n^{-2}) \) and \( a_n:=\int_{1}^{\pi n}\frac{(\sin (5x))^2}{x}\,d x \), then by Stolz-Cesáro we have that
\( \displaystyle{
\lim_{n\to \infty }\frac{a_n}{b_n}=\lim_{n\to \infty }\frac{a_n-a_{n-1}}{b_n-b_{n-1}}
} \)
as far as the later limit exists. Now observe that
\( \displaystyle{
a_n-a_{n-1}=\int_{(n-1)\pi}^{n\pi}\frac{(\sin (5x))^2}{x}\,d x\in \left[\frac{C}{\pi n},\frac{C}{\pi(n-1)}\right],\quad \text{ where }C:=\int_{0}^\pi(\sin (5x))^2 \,d x
} \)
and
\( \displaystyle{
b_n-b_{n-1}=\ln\left(\frac{(n^4+1)(n-1)^2}{((n-1)^4+1)n^2}\right)\implies \lim_{n\to \infty }(n+t)\ln\left(\frac{(n^4+1)(n-1)^2}{((n-1)^4+1)n^2}\right)=2\quad \text{ for any chosen }t\in \mathbb{R}
} \)
Therefore our limit is \( \frac{C}{2\pi}=\frac1{4} \) (as it can be checked easily that \( C=\pi/2 \)).
Addition: to handle the limit with the logarithm easily you can write
\( \displaystyle{
\frac{(n^4+1)(n-1)^2}{((n-1)^4+1)n^2}=1+\frac{(n^4+1)(n-1)^2-((n-1)^4+1)n^2}{((n-1)^4+1)n^2}=1+\frac2{n}\left(\frac{1+\frac{o(n^5)}{n^5}}{1+\frac{o(n^5)}{n^5}}\right)
} \)
Second addition: if the original limit is a functional limit (that is, \( n\in \mathbb{R} \)) then we will find the same answer, because \( \ln(u^2+u^{-2})\sim_{\infty }2\ln u \) and as \( \sin ^2(5x)=\frac{1-\cos (10x)}{2} \) then for any chosen \( a>0 \) we find that
\( \displaystyle{
\lim_{u\to \infty }\frac1{\ln (u^2+u^{-2})}\int_{a}^{\pi u}\frac{\sin ^2(5x)}{x}\,d x=\lim_{u\to \infty }\frac1{4\ln u}\int_{10a}^{10\pi u}\frac{1-\cos t}{t}\,d t
} \)
using the change of variable \( 10x=t \). However the last limit is easy to see that is \( 1/4 \) as \( \left|\int_{a}^{\infty }\frac{\cos t}{t}\,d x\right|<\infty \) for any chosen \( a>0 \).∎
