Question

$\underset{x\to 0}{\lim }\frac{6^x-3^x-2^x+1}{x^2}$ is equal to

• A. $\left({\log }_{e}2\right){\log }_{e}3$
• B. ${\log }_{e}5$
• C. ${\log }_{e}6$
• D. 0

Answer 1

Answer: (A)

We have,
$\underset{x\to 0}{\lim }\frac{6^x-3^x-2^x+1}{x^2}=\underset{x\to 0}{\lim }\frac{3^x\cdot 2^x-3^x-2^x+1}{x^2}$
$=\underset{x\to 0}{\lim }\frac{3^x\left(2^x-1\right)-1\left(2^x-1\right)}{x^2}$
$=\underset{x\to 0}{\lim }\frac{3^x-1}{x}\times \underset{x\to 0}{\lim }\frac{2^x-1}{x}$
$=\left({\log }_{e}2\right)\left({\log }_{e}3\right)$