Question

$\underset{x\to 0}{lim}\frac{{10}^x-2^x-5^x+1}{x\tan x}$ is equal to

• A. ${\log }_{e}2$
• B. ${\log }_{5}2$
• C. $\left({\log }_{e}2\right)\left({\log }_{e}5\right)$
• D. ${\log }_{e}5$

Answer 1

The correct option is C $\left({\log }_{e}2\right)\left({\log }_{e}5\right)$

$Wehave{lim}_{x\to 0}\frac{{10}^x-2^x-5^x+1}{x\tan x}$

$={lim}_{x\to 0}\frac{2^x\left(5^x-1\right)-1\left(5^x-1\right)}{x\tan x}$

$={lim}_{x\to 0}\frac{\left(2^x-1\right)\left(5^x-1\right)}{x\tan x}=\left(\frac{2^x-1}{x}\right)\times \frac{\left(5^x-1\right)}{\left(\frac{\tan x}{x}\right)x}=\left(\frac{2^x-1}{x}\right)\times \frac{\left(5^x-1\right)}{x}\times \frac{1}{\left(\frac{\tan x}{x}\right)}$

$={\log }_{e}2.{\log }_{e}5\times 1$

$={\log }_{e}2.{\log }_{e}5$

Hence, this is the answer.