integration - Prove $lim_{nrightarrow infty }frac{1}{n}int_{frac{1}{n}}^{1}frac{cos2t}{4t^{2}}mathrm dt=frac{1}{4}$

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.

Answer

It's easy to see that when $0 $$\frac{1}{4t^2}-\frac{1}{2}<\frac{\cos2t}{4t^2}< \frac{1}{4t^2}$$
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.