How to Evaluate $\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$? How to find this limit without using L'Hospital rule
$$\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$$

$Step\,1.$  $(1+x)^a = 1+ax + O(x^2)$ for all a and all $x$ with $|x|<1$
$Step\,2.$  $ 4^x - 3^x= 4^x\left(1-\left(\frac34\right)^x\right)$
$Step\,3.$ $(3/4)^x = \exp(x \ln(3/4)) = 1+ x \ln(3/4)+ \frac{x^2}2 \ln^2(3/4)+ O(x^3)$.
$Step\,4.$ The limit $L= \lim_{x \to 0} \frac{(1-\frac13x + O(x^2)-1)}{4^x ( -x \ln(3/4)+O(x^2))}= \lim_{x \to 0} \frac{-\frac13 x}{ -x\ln3/4 }=-\frac1{3\ln4/3}.$
