<s>[INST] What is the value of $ \lim_{x \to 0} \left [\frac{1}{1 \sin^2 x}+ \frac{1}{2 \sin^2 x} +....+ \frac{1}{n \sin^2 x}\right]^{\sin^2x} $ 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 . 
 [/INST] 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 $$\lim_{x \rightarrow 0}  \Bigg( \frac{1}{\sin^2x}+\frac{1}{2\sin^2x}+...+\frac{1}{n\sin^2x}  \Bigg)^{\sin^2x }=\lim_{x \rightarrow 0} \Big( \frac{1}{\sin^2x} \Big)^{\sin^2x} \Bigg( 1+\frac{1}{2}+...+\frac{1}{n}  \Bigg)^{\sin^2x }=1. $$
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