Question

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

• A. $\infty$
• B. $\frac{1}{3}$
• C. $-\frac{1}{3}$
• D. does not exist

Answer 1

The correct option is C $-\frac{1}{3}$

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

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

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

$=2\times \left(\frac{-1}{6}\right)$ ........ $(\because \underset{x\to 0}{lim}\frac{\sin x-x}{x^3}=\frac{-1}{6})$

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