Question

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

• A. $\log 2$
• B. $\frac{\log 2}{\log 5}$
• C. $\left(\log 2\right)\left(\log 5\right)$
• D. $\log 10$

Answer 1

Answer: (C)

$\left(\log 2\right)\left(\log 5\right)$

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

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

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

$=\left(\log 5\right).\left(\log 2\right)$