How to evaluate the following limit: $\lim_{x\to 0}\frac{12^x-4^x}{9^x-3^x}$?

Yes, you can evaluate the limit without LHospital's rule as follows $$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\lim_{x\to 0}\dfrac{4^x\left(\left(\frac{12}{4}\right)^x-1\right)}{3^x\left(\left(\frac93\right)^x-1\right)}$$ $$=\lim_{x\to 0}\dfrac{4^x\left(3^x-1\right)}{3^x(3^x-1)}$$ $$=\lim_{x\to 0}\left(\dfrac{4}{3}\right)^x$$ $$=\color{blue}{1}$$

A variation of other answers (that more closely parallels a common pattern when the numerator and denominator are polynomials) is "big part factoring". \begin{align*} \lim_{x \rightarrow 0} \frac{12^x - 4^x}{9^x-3^x} &= \lim_{x \rightarrow 0} \frac{12^x \left(1 - \left( \frac{4}{12} \right) ^x \right)}{9^x \left( 1-\left( \frac{3}{9} \right) ^x \right) } \\ &= \lim_{x \rightarrow 0} \frac{12^x \left(1 - \left( \frac{1}{3} \right) ^x \right)}{9^x \left( 1-\left( \frac{1}{3} \right) ^x \right) } \\ &= \lim_{x \rightarrow 0} \frac{12^x }{9^x} \\ &= \frac{1}{1} \\ &= 1 \text{.} \end{align*}

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

$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\frac{\ln(\frac{12}{4})}{\ln(\frac{9}{3})}=\frac{\ln(3)}{\ln(3)}=1$$