Question

${lim}_{x\to 0}(\frac{1}{x^2}-\frac{1}{si{n}^{2}x})$

Answer 1

Consider the given function:

$f\left(x\right)={lim}_{x\to 0}(\frac{1}{x^2}-\frac{1}{{\sin }^{2}x})$

$={lim}_{x\to 0}\left(\frac{{\sin }^{2}x-x^2}{x^2{\sin }^{2}x}\right).\frac{x^2}{x^2}$

$={lim}_{x\to 0}\left(\frac{{\sin }^{2}x-x^2}{x^4}\right)\left(\frac{x^2}{{\sin }^{2}x}\right)$

$weknowthat\left(\frac{0}{0}\right)form,$

$={lim}_{x\to 0}\frac{2\sin x\cos x-2x}{4x^3}.1$

$={lim}_{x\to 0}\frac{2(-\sin x\sin x+\cos x\cos x)-2}{12x^2}$

$={lim}_{x\to 0}\frac{2({\cos }^{2}x-{\sin }^{2}x)-2}{12x^2}$

$={lim}_{x\to 0}\frac{2\cos 2x-2}{12x^2}$

$={lim}_{x\to 0}\frac{-4\sin 2x-0}{24x^2}$

$={lim}_{x\to 0}-4\left(\frac{\sin 2x}{2x}\right).\left(\frac{1}{2}\right)$

$=\frac{-4}{12}.1=-\frac{1}{3}$

Hence this is the answer.