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Q.
Let $f:[0,1] \rightarrow R$ be a differentiable function with $f(0)=0$, then $\underset{n \rightarrow \infty}{\text{Lim}} n^2 \int\limits_0^{\frac{1}{n}} f(t) d t$ equals
Integrals
Solution:
Let $I=\underset{n \rightarrow \infty}{\text{Lim}} n^2 \int\limits_0^{\frac{1}{n}} f(t) d t=\underset{n \rightarrow \infty}{\text{Lim}} \frac{\int\limits_0^{1 / n} f(t) d t}{\frac{1}{n^2}} \quad\left(\frac{0}{0}\right.$ form, as $\left.n \rightarrow \infty\right)$
Using L'Hospital's rule and Using Leibnitz rule, we get
$I=\underset{n \rightarrow \infty}{\text{Lim}}\frac{n^3 f\left(\frac{1}{n}\right)\left(\frac{-1}{n^2}\right)}{-2}=\underset{n \rightarrow \infty}{\text{Lim}}\frac{n}{2} f\left(\frac{1}{n}\right)$
Put $n =\frac{1}{ h }$, hence
$I=\frac{1}{2} \underset{n \rightarrow \infty}{\text{Lim}}\left(\frac{f(0+h)-f(0)}{h}\right)=\frac{1}{2} f^{\prime}(0) $