Question

The value of $\underset{x\to 0}{lim}(1-\frac{1}{2^x})\left(\frac{1}{\sqrt{\tan x+4}-2}\right)$ is:

• A. $\log 16$
• B. does not exist
• C. $3\log 2$
• D. $6\log 2$

Answer 1

Answer: (A)

$\log 16$

$\underset{x\to 0}{lim}(1-\frac{1}{2^x})\left(\frac{1}{\sqrt{\tan x+4-2}}\right)$

$=\underset{x\to 0}{lim}\left(\frac{2^x-1}{2^x}\right)(\frac{1}{\sqrt{\tan x+4}-2}\times \frac{\sqrt{\tan x+4}+2}{\sqrt{\tan x+4}+2})$

$=\underset{x\to 0}{lim}\left(\frac{2^x-1}{2^x}\right)\left(\frac{\sqrt{\tan x+4}+2}{\tan x}\right)$

$=\underset{x\to 0}{lim}\left(\frac{2^x-1}{2^{x}x}\right)\left(\frac{\sqrt{\tan x+4}+2}{\left(\frac{\tan x}{x}\right)}\right)$
Applying L'Hospital's rule, in first limit

$=\underset{x\to 0}{lim}\left(\frac{2^x\log 2}{2^{x}x\log 2+2^x}\right)\left(\frac{\sqrt{\tan x+4}+2}{\left(\frac{\tan x}{x}\right)}\right)$

$\Rightarrow 4\log 2=\log 16$