<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] As an alternative, you can use
$$a^{x} = e^{x \ln(a)} = 1 + x\ln(a) + \frac{x^2\ln^{2}(a)}{2!}  + \mathcal{O}(x^{3})$$
therefore
\begin{align}
\frac{a^{x} - b^{x}}{c^{x} - d^{x}} &= \frac{x\,(\ln(a) - \ln(b)) + \frac{x^2}{2} \, (\ln^{2}(a) - \ln^{2}(b))  + \mathcal{O}(x^{3})}{x\,(\ln(c) - \ln(d)) + \frac{x^2}{2} \, (\ln^{2}(c) - \ln^{2}(d)) + \mathcal{O}(x^{3})} \\
&= \frac{\ln(\frac{a}{b}) + \frac{x}{2} \, \ln(a b)\,\ln(\frac{a}{b})   + \mathcal{O}(x^{2})}{\ln(\frac{c}{d}) + \frac{x}{2} \, \ln(c d)\,\ln(\frac{c}{d})   + \mathcal{O}(x^{2})}
\end{align}
Taking the limit as $x \to 0$ gives
$$\lim_{x \to 0}  \frac{a^{x} - b^{x}}{c^{x} - d^{x}} = \frac{\ln(\frac{a}{b})}{\ln(\frac{c}{d})} $$
hence for your limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\frac{\ln(\frac{12}{4})}{\ln(\frac{9}{3})}=\frac{\ln(3)}{\ln(3)}=1$$
</s>