$\displaystyle\lim_{x\rightarrow 0}\left(\dfrac{1}{\sin^2x}-\dfrac{1}{\sin h^2x}\right)=?$

$\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{1}{\sin ^{2} x}-\frac{1}{\sinh ^{2} x}\right) \\ = & \lim _{x \rightarrow 0}\left(\frac{\sinh ^{2} x-\sin ^{2} x}{\sin ^{2} x \sinh ^{2} x}\right)\end{aligned}$

$=\lim _{x \rightarrow 0} \frac{\left(x+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\cdots \right)^{2}-\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\cdots\right)^{2}}{\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots\right)^{2}\left(x+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+…\right)^{2}}$

$=\lim _{x \rightarrow 0} \frac{\left(x^{2}+\frac{x^{6}}{(3!)^{2}}+\frac{2 x^{4}}{3!}\right)-\left(x^{2}+\frac{x^{6}}{(3!)^{2}}-\frac{2 x^{4}}{3!}\right)}{x^{4}\left(1-\frac{x^{2}}{(3!)^{2}}-\cdots\right)\left(1+\frac{x^{2}}{(3)^{2}}+\cdots\right)}$

$=\lim _{x \rightarrow 0} \frac{2 \cdot \frac{2 x^{4}}{3!}}{x^{4} \cdot 1}$

$=\frac{4}{6} =\frac{2}{3}$