Question

$\underset{x\to 0}{\lim }\left(\frac{\ln \cos x}{\sqrt[4]{1+x^2}-1}\right)$ is equal to

• A. 2
• B. -2
• C. 1
• D. -1

Answer 1

Answer: (B)

$\underset{x\to 0}{\lim }\left(\frac{\ln \cos x}{\sqrt[4]{1+x^2}-1}\right)$
$=\underset{x\to 0}{\lim }\frac{\ln [1+(\cos x-1)]}{\sqrt[4]{1+x^2}-1}$
$=4\underset{x\to 0}{\lim }\frac{\cos x-1}{x^2}$
$[\because \ln [1+(\cos x-1)]\sim (\cos x-1)$ and $\left(\sqrt[4]{1+x^2}-1\right)\sim \frac{x^2}{4}]$
$=-4\underset{x\to 0}{\lim }\frac{x^2/2}{x^2}$
$\left[\because (1-\cos x)\sim \frac{x^2}{2}\right]$
$=-2$