Question

$\underset{x\to 0}{lim}\frac{\sqrt{\cos x}-\sqrt[3]{3\cos x}}{{\sin }^{2}x}$=

• A. $\frac{1}{3}$
• B. $-\frac{1}{3}$
• C. $\frac{1}{12}$
• D. $-\frac{1}{12}$

Answer 1

Answer: (D)

$-\frac{1}{12}$

$\underset{x\to 0}{lim}\frac{\sqrt{\cos x}-\sqrt[3]{3\cos x}}{{\sin }^{2}x}$

It is of the form $\frac{0}{0}$, so using L-Hospital's rule,
$=\underset{x\to 0}{lim}\frac{\frac{-\sin x}{2\sqrt{\cos x}}-\frac{-\sin x{\cos }^{\frac{-2}{3}}x}{3}}{2\sin x\cos x}$

$=-\frac{1}{12}$