I took out $\sin^2x$ out of the brackets . Inside the brackets , I think I should use the formula $\frac{ n(n-1)}{2}$ . Am I doing right ? If yes, then what should I do next ? Thanks !
I am sorry for the heading . I know it is not clear and I have no idea why it is not being displayed properly .
You have the limit is $1$ because:
$1)$: $\displaystyle \lim_{x\to 0}\left(\displaystyle \sum_{j=1}^{n} \dfrac{1}{j}\right)^{\sin^2x}=1$,
$2)$: $\displaystyle \lim_{x\to 0} \left(\dfrac{1}{\sin^2x}\right)^{\sin^2x} = 1$.
we have $$ \Big( \frac{1}{\sin^2x} \Big)^{\sin^2x} = e^{\sin^2x \ln \frac{1}{\sin^2x}}$$knowing that ${\sin^2x} \rightarrow 0$ as $x \rightarrow 0$, and $ \lim_{x\rightarrow 0} x\ln \frac{1}{x} = 0 $ then the limit is $$\lim_{x \rightarrow 0} \Big( \frac{1}{\sin^2x} \Big)^{\sin^2x} =1. $$ Similarly, $$\lim_{x \rightarrow 0} \Bigg( 1+\frac{1}{2}+...+\frac{1}{n} \Bigg)^{\sin^2x } = \lim_{x \rightarrow 0} e^{\sin^2x \ln{( 1+\frac{1}{2}+...+\frac{1}{n} )} }=1. $$ SO $$\lim_{x \rightarrow 0} \Bigg( \frac{1}{\sin^2x}+\frac{1}{2\sin^2x}+...+\frac{1}{n\sin^2x} \Bigg)^{\sin^2x }=\lim_{x \rightarrow 0} \Big( \frac{1}{\sin^2x} \Big)^{\sin^2x} \Bigg( 1+\frac{1}{2}+...+\frac{1}{n} \Bigg)^{\sin^2x }=1. $$
