Question

What is the value of $\underset{x\to 0}{lim}(\frac{1}{x^2}-\frac{1}{{\sin }^{2}x})$?

Answer 1

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

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

Multiply and divide by $x^2$ in denominator, we have

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

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

$\because {lim}_{x\to 0}\frac{\sin x}{x}=1$

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

$\Rightarrow ={lim}_{x\to 0}\frac{(x-\sin x)(x+\sin x)}{x^4}$

$\Rightarrow =\left({lim}_{x\to 0}\frac{x-\sin x}{x^3}\right)\left({lim}_{x\to 0}\frac{x+\sin x}{x}\right)$

$\Rightarrow ={lim}_{x\to 0}\left[\left(\frac{x-\sin x}{x^3}\right)\left(\frac{x+\sin x}{x}\right)\right]$

Applying L'Hospital rule, we have

${lim}_{x\to 0}(\frac{1}{x^2}-\frac{1}{{\sin }^{2}x})=\left({lim}_{x\to 0}\frac{1+\cos x}{1}\right)\left({lim}_{x\to 0}\frac{1-\cos x}{3x^2}\right)$

$\Rightarrow =\left(\frac{1+1}{1}\right)\left({lim}_{x\to 0}\frac{1-\cos x}{3x^2}\right)$

Again applying L'Hospital rule, we have

$\Rightarrow =2\left({lim}_{x\to 0}\frac{\sin x}{6x}\right)$

$\Rightarrow =2\times \frac{1}{6}[\because {lim}_{x\to 0}\frac{\sin x}{x}=1]$

$\Rightarrow =\frac{1}{3}$

Hence ${lim}_{x\to 0}(\frac{1}{x^2}-\frac{1}{{\sin }^{2}x})=\frac{1}{3}$.