Evaluate $\lim\limits_{x \to 0} \; \dfrac{4^x - 9^x}{x \left(4^x + 9^x\right)}$
$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{4^x - 9^x}{x \left(4^x + 9^x\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(4^x - 9^x\right)}{\dfrac{d}{dx}\left[x\left(4^x + 9^x\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{4^x \ln 4 - 9^x \ln 9}{4^x + 9^x + x \left(4^x \ln 4 + 9^x \ln 9\right)} \\ & & \\ & = \dfrac{\ln 4 - \ln 9}{1+1+0} \\ & & \\ & = \dfrac{1}{2} \ln \left(\dfrac{4}{9}\right) \\ & & \\ & = \dfrac{1}{2} \ln \left(\dfrac{2}{3}\right)^2 \\ & & \\ & = \dfrac{1}{2} \times 2 \ln \left(\dfrac{2}{3}\right) = \ln \left(\dfrac{2}{3}\right) \end{aligned}$