<s>[INST] How to evaluate the following limit: $\lim_{x\to 0}\frac{12^x-4^x}{9^x-3^x}$? How can I compute this limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}\text{?}$$
My solution is here:
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\dfrac{1-1}{1-1} = \dfrac{0}{0}$$
I used L'H$\hat{\mathrm{o}}$pital's rule:
\begin{align*}
\lim_{x\to 0}\dfrac{12^x\ln12-4^x\ln4}{9^x\ln9-3^x\ln3}&=\dfrac{\ln12-\ln4}{\ln9-\ln3}
\\ &=\dfrac{\ln(12/4)}{\ln(9/3)}
\\ &=\dfrac{\ln(3)}{\ln(3)}
\\ &=1
\end{align*}
My answer comes out to be $1$. Can I evaluate this limit without L'H$\hat{\mathrm{o}}$pital's rule? Thanks.
 [/INST] Yes, you can evaluate the limit without LHospital's rule as follows
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\lim_{x\to 0}\dfrac{4^x\left(\left(\frac{12}{4}\right)^x-1\right)}{3^x\left(\left(\frac93\right)^x-1\right)}$$
$$=\lim_{x\to 0}\dfrac{4^x\left(3^x-1\right)}{3^x(3^x-1)}$$
$$=\lim_{x\to 0}\left(\dfrac{4}{3}\right)^x$$
$$=\color{blue}{1}$$
</s>