Question

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

• A. ${\log }_e\frac{3}{2}$
• B. ${\log }_{e}5$
• C. ${\log }_{e}6$
• D. ${\log }_{e}2{\log }_{e}3$

Answer 1

Answer: (D)

$\underset{x\to 0}{\lim }\frac{6^x-3^x-2^x+1}{x^2}$
Applying L-Hospital's rule,
$=\underset{x\to 0}{\lim }\frac{6^x{\log }_{e}6-3^x{\log }_{e}3-2^x{\log }_{e}2}{2x}$
Again, applying L-Hospital's rule,
$\underset{x\to 0}{\lim }\frac{6^x{\left({\log }_{e}6\right)}^2-3^x{\left({\log }_{e}3\right)}^2-2^x{\left({\log }_{e}2\right)}^2}{2}$
$=\frac{{\left({\log }_{e}6\right)}^2-{\left({\log }_{e}3\right)}^2-{\left({\log }_{e}2\right)}^2}{2}$
$=\frac{{\left({\log }_{e}3.2\right)}^2-{\left({\log }_{e}3\right)}^2-{\left({\log }_{e}2\right)}^2}{2}$
$=\frac{{\left({\log }_{e}3+{\log }_{e}2\right)}^2-{\left({\log }_{e}3\right)}^2-{\left({\log }_{e}2\right)}^2}{2}$
$=\frac{{\left({\log }_{e}3\right)}^2-{\left({\log }_{e}2\right)}^2+2{\log }_{e}3{\log }_{e}2-{\left({\log }_{e}3\right)}^2-{\left({\log }_{e}2\right)}^2}{2}$
$={\log }_{e}3{\log }_{e}2$