Question

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

• A. $\frac{1}{3}$
• B. $-\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $-\frac{1}{6}$

Answer 1

Answer: (B)

$-\frac{1}{3}$

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

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

Apply L-Hospital's rule

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

Apply L-Hospital's rule

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

Apply L-Hospital's rule

$={lim}_{x\to 0}\left(\frac{-2\sin \left(2x\right)}{-2x^2\sin \left(2x\right)+6x\cos \left(2x\right)+3\sin \left(2x\right)}\right)$

Apply L-Hospital's rule

$={lim}_{x\to 0}(-\frac{\cos \left(2x\right)}{-x^2\cos \left(2x\right)-4x\sin \left(2x\right)+3\cos \left(2x\right)})$

$=-\frac{\cos (2\cdot 0)}{-0^2\cos (2\cdot 0)-4\cdot 0\cdot \sin (2\cdot 0)+3\cos (2\cdot 0)}$

upon simplification, we get

$-\frac{1}{3}$