$\lim _{x \rightarrow 0} \frac{9^x-4^x}{x\left(9^x+4^x\right)}=$
$\lim _{x \rightarrow 0} \frac{9^x-4^x}{x\left(9^x+4^x\right)}=$
- $\quad \log \left(\frac{3}{2}\right)$
- $\frac{1}{2} \log \left(\frac{3}{2}\right)$
- $\quad 2 \log \left(\frac{3}{2}\right)$
- $2 \log \left(\frac{9}{4}\right)$
Solution
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{9^x-4^x}{x\left(9^x+4^x\right)} \\ & =\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)-\left(4^x-1\right)}{x} \times \frac{1}{\left(9^x+4^x\right)} \\ & =\left[\lim _{x \rightarrow 0} \frac{9^x-1}{x}-\lim _{x \rightarrow 0} \frac{4^x-1}{x}\right] \times \lim _{x \rightarrow 0} \frac{1}{\left(9^x \cdot+4^x\right)} \\ & =\log \left(\frac{9}{4}\right) \times \frac{1}{(1+1)} \\ & =\log \left(\frac{3}{2}\right)\end{aligned}$
Asked in: MHT CET 2024 (10 May Shift 2)
Practice more Limits questions on Aicharya