Prove $\lim_{n\rightarrow \infty }\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\mathrm dt=\frac{1}{4}$ How to prove
$$\lim_{n\rightarrow \infty }\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\mathrm dt=\frac{1}{4}$$
I used $x\to \dfrac{1}{t}$ but it didn't work.
Any hint?Thank you.
 A: It's easy to see that when $0<t\leq 1$, we have $1-2t^2< \cos2t<1$, hence
$$\frac{1}{4t^2}-\frac{1}{2}<\frac{\cos2t}{4t^2}< \frac{1}{4t^2}$$
so
$$\frac{1}{n}\int_{\frac{1}{n}}^{1}\left ( \frac{1}{4t^2}-\frac{1}{2} \right )\mathrm{d}t<\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\, \mathrm{d}t<\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{1}{4t^2}\, \mathrm{d}t$$
$$\Rightarrow \frac{1}{4}-\frac{3}{4n}+\frac{1}{2n^2}<\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\, \mathrm{d}t<\frac{1}{4}-\frac{1}{4n}$$
Now take the limit we will get the answer as wanted.
A: METHODOLOGY $1$: L'Hospital's. Rule
One simple and efficient approach is to use L'Hospital"s Rule ( See Note at the end of this Section ). 
Note that we have
$$\begin{align}
\lim_{n\to \infty}\frac{\int_{1/n}^1\frac{\cos(2t)}{4t^2}\,dt}{n}&=\lim_{n\to \infty}\frac{d}{dn}\int_{1/n}^1\frac{\cos(2t)}{4t^2}\,dt\\\\
&=\lim_{n\to \infty}\left(-\frac{\cos(2/n)}{4/n^2}\times \frac{-1}{n^2}\right)\\\\
&=\frac14
\end{align}$$
as expected!

METHODOLOGY $2$: Change of Variable and the Dominated Convergence Theorem
Enforcing the substitution $t=x/n$ reveals
$$\begin{align}
\lim_{n\to \infty}\frac1n \int_{1/n}^1\frac{\cos(2t)}{4t^2}\,dt&=\lim_{n\to \infty}\int_1^n \frac{\cos(2x/n)}{4x^2} \,dx\\\\ 
&=\lim_{n\to \infty}\int_1^\infty \xi_{[0,n]}\frac{\cos(2x/n)}{4x^2}\,dx\\\\
&=\int_1^\infty \lim_{n\to \infty}\xi_{[0,n]}\frac{\cos(2x/n)}{4x^2}\,dx\\\\
&=\int_1^\infty \frac1{4x^2}\,dx\\\\
&=\frac14
\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\lim_{n \to \infty }\bracks{{1 \over n}
\int_{1/n}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t} = {1 \over 4}:\ {\large ?}}$.

With Stolz-Ces$\mathrm{\grave{a}}$ro Theorem:
\begin{align}
&\lim_{n \to \infty }\bracks{{1 \over n}
\int_{1/n}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t} =
\lim_{n \to \infty }\braces{{1 \over \pars{n + 1} - n}
\bracks{\int_{1/\pars{n + 1}}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t -
\int_{1/n}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t}}
\\[5mm] = &\
\lim_{n \to \infty }\int_{1/\pars{n + 1}}^{1/n}{\cos\pars{2t} \over 4t^{2}}
\,\dd t =
\lim_{n \to \infty}\bracks{\pars{{1 \over n} - {1 \over n + 1}}{\cos\pars{2\xi_{n}} \over 4\xi_{n}^{2}}}
\quad\mbox{where}\quad {1 \over n + 1} < \xi_{n} < {1 \over n}
\end{align}

Moreover $\ds{\pars{~\mbox{note that}\ \lim_{n \to \infty}\,\,\xi_{n} = 0~}}$,
\begin{align}
&\lim_{n \to \infty }\bracks{{1 \over n}
\int_{1/n}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t} =
{1 \over 4}\lim_{n \to \infty}
\bracks{{1 \over n\pars{n + 1}}{\cos\pars{2\xi_{n}} \over \xi_{n}^{2}}} =
{1 \over 4}\lim_{n \to \infty}\,{1 \over n\pars{n + 1}\xi_{n}^{2}}\label{1}\tag{1}
\end{align}


Note that
  $\ds{{n \over n + 1} < n\pars{n + 1}\xi_{n}^{2} < {n + 1 \over n}}$ such that
  $\ds{\lim_{n \to \infty}\,{1 \over n\pars{n + 1}\xi_{n}^{2}} = 1}$.

From expression \eqref{1}:
$\ds{\lim_{n \to \infty }\bracks{{1 \over n}
\int_{1/n}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t} = \bbx{\ds{1 \over 4}}}$.
A: For reference, here is a solution based on integration by parts:
\begin{align*}
\int_{1/n}^{1} \frac{\cos 2t}{4t^2} \, dt
&= \left[ - \frac{\cos 2t}{4t} \right]_{1/n}^{1} - \int_{1/n}^{1} \frac{\sin 2t}{2t} \, dt \\
&= \frac{n}{4}\cos(2/n) - \frac{\cos2}{4} - \int_{1/n}^{1} \frac{\sin 2t}{2t} \, dt.
\end{align*}
The last integral term causes no harm, since $|\sin 2t| \leq |2t|$ and hence $ \left| \int_{1/n}^{1} \frac{\sin 2t}{2t} \, dt \right| \leq 1$. Thus by the squeezing theorem the claim follows.
