Question

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

• A. 1
• B. -1
• C. 1/2
• D. -1/2

Answer 1

Answer: (D)

$\underset{x\to 0}{\lim }\frac{\cos (\sin x)-1}{x^2}$
Now, using L-Hopital's rule,
$=\underset{x\to 0}{\lim }\frac{-\sin (\sin x)\cdot \cos x}{2x}$
Again using L-Hopital's rule
$=\underset{x\to 0}{\lim }\frac{-[-\sin (\sin x)\times \sin x+{\cos }^{2}x}{\times \cos (\sin x)]}$
$=-\frac{1}{2}$