Question

$\lim_{x \rightarrow 0}\frac{4^{x} - 9^{x}}{x(4^{x} + 9^{x})} =$

• A. $\log\left( \frac{2}{3} \right)$
• B. $\frac{1}{2}\log\left( \frac{3}{2} \right)$
• C. $\frac{1}{2}\log\left( \frac{3}{2} \right)$
• D. $\log\left( \frac{3}{2} \right)$

Answer 1

Answer: (A)

$\log\left( \frac{2}{3} \right)$

Answer 2

$y = \lim _ { x \rightarrow 0 } \frac { 4 ^ { x } - 9 ^ { x } } { x \left( 4 ^ { x } + 9 ^ { x } \right) }$ $\left( \frac{0}{0}\text{form} \right)$

Using L-Hospital’s rule,

$y = \lim_{x \rightarrow 0}\frac{4^{x}\log 4 - 9^{x}\log 9}{(4^{x} + 9^{x}) + x(4^{x}\log 4 + 9^{x}\log 9)}$⇒$y = \frac{\log 4 - \log 9}{2}$ ⇒ $y = \frac{{\log\left( \frac{2}{3} \right)}^{2}}{2} = \log\frac{2}{3}$.