$\lim _{x \rightarrow 0} \frac{4^x-9^x}{x\left(4^x+9^x\right)}$ is equal to
$\lim _{x \rightarrow 0} \frac{4^x-9^x}{x\left(4^x+9^x\right)}$ is equal to
- $\log \frac{2}{3}$
- $\log \frac{3}{2}$
- $\frac{1}{2} \log \frac{2}{3}$
- $\frac{1}{2} \log \frac{3}{2}$
Solution
We have,
$
\lim _{x \rightarrow 0} \frac{4^x-9^x}{x\left(4^x+9^x\right)}
$
Using L-Hospital's rule
$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{4^x \log 4-9^x \log 9}{\left(4^x+9^x\right)+x\left(4^x \log 4+9^x \log 9\right)} \\
& =\frac{\log 4-\log 9}{2}=\frac{2 \log \frac{2}{3}}{2}=\log \frac{2}{3}
\end{aligned}
$
Asked in: AP EAMCET 2002
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