$\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\left(\frac{1}{{\sin }^2 \mathrm{x}}-\frac{1}{{\sinh }^2}\right)=$

$\begin{array}{l}\underset{x\to 0}{Lt}\left(\frac{1}{{\sin }^{2}x}-\frac{1}{\sin h^{2}x}\right)\\ =\underset{x\to 0}{Lt}\frac{{\sinh }^{2}x-{\sin }^{2}x}{{\sin }^{2}x{\sinh }^{2}x}\\ =\underset{x\to 0}{lt}\frac{{\sinh }^{2}x-{\sin }^{2}x}{x^4}.\frac{x^2}{{\sin }^{2}x}.\frac{x^2}{{\sinh }^{2}x}\\ =\underset{x\to 0}{Lt}\frac{{\sinh }^{2}x-{\sin }^{2}x}{x^4}\left(\frac{0}{0}Form\right)\\ =\underset{x\to 0}{Lt}\frac{\sinh 2x-\sin 2x}{4x^3}\left(\text{L'}hospital\right)\\ =\underset{x\to 0}{Lt}\frac{2\cosh 2x-2\cos 2x}{12x^2}\left(\frac{0}{0}formL\text{'}hospital\right)\\ =\underset{x\to 0}{Lt}\frac{4\sinh 2x+4\sin 2x}{24x}\\ =\underset{x\to 0}{Lt}\frac{1}{6}\left(\frac{\sinh 2x}{x}+\frac{\sin 2x}{x}\right)\\ =\frac{1}{6}\left(2+2\right)=\frac{2}{3}\end{array}$