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$\lim _{x \rightarrow 0} \frac{4^x-9^x}{x\left(4^x+9^x\right)}$ is equal to
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1408 Upvotes
Verified Answer
The correct answer is:
$\log \frac{2}{3}$
We have,
$$
\lim _{x \rightarrow 0} \frac{4^x-9^x}{x\left(4^x+9^x\right)}
$$
Using L-Hospital's rule
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{4^x \log 4-9^x \log 9}{\left(4^x+9^x\right)+x\left(4^x \log 4+9^x \log 9\right)} \\
& =\frac{\log 4-\log 9}{2}=\frac{2 \log \frac{2}{3}}{2}=\log \frac{2}{3}
\end{aligned}
$$
$$
\lim _{x \rightarrow 0} \frac{4^x-9^x}{x\left(4^x+9^x\right)}
$$
Using L-Hospital's rule
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{4^x \log 4-9^x \log 9}{\left(4^x+9^x\right)+x\left(4^x \log 4+9^x \log 9\right)} \\
& =\frac{\log 4-\log 9}{2}=\frac{2 \log \frac{2}{3}}{2}=\log \frac{2}{3}
\end{aligned}
$$
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