Question

The value of $\underset{x\to 0}{lim}\frac{1-4^x-5^x+{20}^x}{\sqrt{2\cos x+7}-3}$ is

• A. $3\ln 4\ln 25$
• B. $3\ln 4\ln 5$
• C. $-3\ln 4\ln 25$
• D. $-3\ln 4\ln 5$

Answer 1

The correct option is C $-3\ln 4\ln 25$

$\underset{x\to 0}{lim}\frac{1-4^x-5^x+{20}^x}{\sqrt{2\cos x+7}-3}$
$=\underset{x\to 0}{lim}\frac{(4^x-1)(5^x-1)(\sqrt{2\cos x+7}+3)}{(2\cos x+7)-9}$
$=\underset{x\to 0}{lim}\frac{\left(\frac{4^x-1}{x}\right)\left(\frac{5^x-1}{x}\right)(\sqrt{2\cos x+7}+3)}{\frac{-2(1-\cos x)}{x^2}}$
$=\underset{x\to 0}{lim}\frac{\left(\frac{4^x-1}{x}\right)\left(\frac{5^x-1}{x}\right)(\sqrt{2\cos x+7}+3)}{\frac{-{\sin }^2\frac{x}{2}}{\frac{x^2}{4}}}$
$=-6\ln 4\ln 5$
$=-3\ln 4\ln 25$