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.
 A: 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*}
A: 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
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\frac{\ln(\frac{12}{4})}{\ln(\frac{9}{3})}=\frac{\ln(3)}{\ln(3)}=1$$
A: It could be amazing to get more than the limit itself for the most general case. Replacing $t^x$ by $e^{x \log(t)}$, using Taylor and long division would give
$$\frac{a^x-b^x}{c^x-d^x}=\frac{\log \left(\frac{a}{b}\right)}{\log \left(\frac{c}{d}\right)}+\frac 12\frac{\log \left(\frac{a}{b}\right) \log \left(\frac{a b}{c d}\right)}{\log
   \left(\frac{c}{d}\right)}x+O(x^2)$$ which shows the limit and also hoaw it is approached.
A: 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
$$
\lim_{x\to0}\frac{\dfrac{12^x-1}{x}-\dfrac{4^x-1}{x}}{\dfrac{9^x-1}{x}-\dfrac{3^x-1}{x}}=\frac{\log12-\log4}{\log9-\log3}=\frac{\log3}{\log3}=1
$$
A: 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: You can also avoid using L'Hopital's rule by the properties of logarithms:
$$\lim_{x\to 0} \exp \left( \ln\dfrac{12^x-4^x}{9^x-3^x} \right)$$
$$= \lim_{x\to 0} \exp \left( \ln (12^x-4^x) - \ln(9^x-3^x) \right)$$
$$= \lim_{x\to 0} \exp \left( (\ln 4^x + \ln(3^x - 1))- (\ln3^x + \ln(3^x - 1)) \right)$$
$$= \lim_{x\to 0} \exp \left(\ln(4^x) - \ln(3^x) \right)$$
$$= \lim_{x\to 0} \exp \left(x \ln 4 - x \ln 3 \right)$$
$$=  \exp(0) = \boxed{1}.$$
See, no fractions! Except for the first line which is a rephrasing of the problem statement, of course.
