Question

Evaluate:
$\underset{x\to 0}{lim}\frac{1}{x^2}-\frac{1}{{\sin }^{2}x}$

Answer 1

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

$\underset{x\to 0}{lim}\left(\frac{{\sin }^{2}x-x^2}{x^2\left(\frac{{\sin }^{2}x}{x^2}\right)\times x^2}\right)[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

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

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

$\underset{x\to 0}{lim}(\frac{\sin x}{x}+1)$ Use L hospitals rule

$1+1=2$

$2\underset{x\to 0}{lim}\left(\frac{\sin x-x}{x^3}\right)$

$2\underset{x\to 0}{lim}\left(\frac{\cos x-1}{3x^2}\right)$

$2\underset{x\to 0}{lim}\left(\frac{-\sin x}{6x}\right)\to 2\underset{x\to 0}{lim}\left(\frac{-\cos x}{6x}\right)=-\frac{1}{6}\times 2$

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