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

If you add and subtract $1$ from numerator and denominator $$ lim_{xto0}frac{(12^x-1)-(4^x-1)}{(9^x-1)-(3^x-1)} $$ then dividing each term by $x$ $$ lim_{xto0}frac{dfrac{12^x-1}{x}-dfrac{4^x-1}{x}}{dfrac{9^x-1}{x}-dfrac{3^x-1}{x}} $$ Now, all four limits have the form $$ lim_{xto0}frac{a^x-1}{x}=log a $$ so we get $$ lim_{xto0}frac{dfrac{12^x-1}{x}-dfrac{4^x-1}{x}}{dfrac{9^x-1}{x}-dfrac{3^x-1}{x}}=frac{log12-log4}{log9-log3}=frac{log3}{log3}=1 $$

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 $$frac{a^x-b^x}{c^x-d^x}=frac{log left(frac{a}{b}right)}{log left(frac{c}{d}right)}+frac 12frac{log left(frac{a}{b}right) log left(frac{a b}{c d}right)}{log left(frac{c}{d}right)}x+O(x^2)$$ which shows the limit and also hoaw it is approached.

You can also avoid using L'Hopital's rule by the properties of logarithms:

$$lim_{xto 0} exp left( lndfrac{12^x-4^x}{9^x-3^x} right)$$ $$= lim_{xto 0} exp left( ln (12^x-4^x) - ln(9^x-3^x) right)$$ $$= lim_{xto 0} exp left( (ln 4^x + ln(3^x - 1))- (ln3^x + ln(3^x - 1)) right)$$ $$= lim_{xto 0} exp left(ln(4^x) - ln(3^x) right)$$ $$= lim_{xto 0} exp left(x ln 4 - x ln 3 right)$$ $$= exp(0) = boxed{1}.$$

See, no fractions! Except for the first line which is a rephrasing of the problem statement, of course.

As an alternative, you can use $$a^{x} = e^{x ln(a)} = 1 + xln(a) + frac{x^2ln^{2}(a)}{2!} + mathcal{O}(x^{3})$$ 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

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

A 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*}

Yes, you can evaluate the limit without LHospital's rule as follows $$lim_{xto 0}dfrac{12^x-4^x}{9^x-3^x}=lim_{xto 0}dfrac{4^xleft(left(frac{12}{4}right)^x-1right)}{3^xleft(left(frac93right)^x-1right)}$$ $$=lim_{xto 0}dfrac{4^xleft(3^x-1right)}{3^x(3^x-1)}$$ $$=lim_{xto 0}left(dfrac{4}{3}right)^x$$ $$=color{blue}{1}$$