Question

$\displaystyle\lim _{x \rightarrow 0^{+}} \log _{\tan x}(\tan 2 x)$ is equal to

• A. 1
• B. -1
• C. 0
• D. None of these

Answer 1

Answer: (A)

$\displaystyle\lim _{x \rightarrow 0^{+}} \log _{\tan x}(\tan 2 x)$
$=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{\log (\tan 2 x)}{\log (\tan x)}\left(\frac{\infty}{\infty}\right.$ form $)$
$=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{\tan 2 x}\left(2 \sec ^{2} 2 x\right)}{\frac{1}{\tan x}\left(\sec ^{2} x\right)}$
[Using L' Hospital's rule]
$=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{2 \sin x \cos x}{\sin 2 x \cos 2 x}=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{2 \sin 2 x}{\sin 4 x}\left(\frac{0}{0}\right.$ form $)$ $=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{4 \cos 2 x}{4 \cos 4 x}=1.$
[Using L' Hospital's rule]