Not sure why my previous answer was deleted...If you substitute [tex]y=\frac1{\sin^2x}=\csc^2x[/tex], then [tex]x\to0[/tex] gives [tex]y\to+\infty[/tex], so the limit is equal in value to[tex]\displaystyle\lim_{y\to\infty}\left(1^y+2^y+\cdots+n^y\right)^{1/y}[/tex]Since [tex]e^{\ln x}=x[/tex], we can write[tex]\left(1^y+2^y+\cdots+n^y\right)^{1/y}=e^{\ln(1^y+2^y+\cdots+n^y)^{1/y}=e^{\ln(1^y+2^y+\cdots+n^y)/y}[/tex]Because [tex]e^x[/tex] is continuous at all [tex]x[/tex], we can pass the limit to the argument of the exponential function. That is,[tex]\displaystyle\lim_{x\to\infty}e^{f(x)}=e^{\lim\limits_{x\to\infty}f(x)}[/tex]so that the limit we're interested in is equal to[tex]e^{\lim\limits_{y\to\infty}\ln(1^y+2^y+\cdots+n^y)/y}[/tex]For all natural numbers [tex]n[/tex], we have[tex]1^y+2^y+\cdots+n^y\le n^y+n^y+\cdots+n^y=n\cdot n^y=n^{y+1}[/tex]so[tex]\displaystyle\lim_{y\to\infty}\frac{\ln(1^y+2^y+\cdots+n^y)}y=\lim_{y\to\infty}\frac{\ln n^{y+1}}y=\ln n\cdot\lim_{y\to\infty}\frac{y+1}y=\ln n[/tex]and this makes the overall limit take on a value of[tex]e^{\ln n}=\boxed n[/tex]