<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] If you add and subtract $1$ from numerator and denominator
$$
\lim_{x\to0}\frac{(12^x-1)-(4^x-1)}{(9^x-1)-(3^x-1)}
$$
then dividing each term by $x$
$$
\lim_{x\to0}\frac{\dfrac{12^x-1}{x}-\dfrac{4^x-1}{x}}{\dfrac{9^x-1}{x}-\dfrac{3^x-1}{x}}
$$
Now, all four limits have the form
$$
\lim_{x\to0}\frac{a^x-1}{x}=\log a
$$
so we get
$$
\lim_{x\to0}\frac{\dfrac{12^x-1}{x}-\dfrac{4^x-1}{x}}{\dfrac{9^x-1}{x}-\dfrac{3^x-1}{x}}=\frac{\log12-\log4}{\log9-\log3}=\frac{\log3}{\log3}=1
$$
</s>