Question

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

• A. 2
• B. -2
• C. 1
• D. -1

Answer 1

Answer: (B)

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