How to Evaluate $\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$?

Question

How to find this limit without using L'Hospital rule

$$\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$$

Answer 1

$Step\,1.$ $(1+x)^a = 1+ax + O(x^2)$ 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 $L= \lim_{x \to 0} \frac{(1-\frac13x + O(x^2)-1)}{4^x ( -x \ln(3/4)+O(x^2))}= \lim_{x \to 0} \frac{-\frac13 x}{ -x\ln3/4 }=-\frac1{3\ln4/3}.$

Answer 2

Use, binomial expansion of $(1-x)^{1/3}$ & Taylor's series expansion of $4^x$ & $3^x$ as follows $$\lim_{x\to 0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$$ $$=\lim_{x\to 0}\frac{\left(1+\frac{\frac{1}{3}}{1!}(-x)+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!}(-x)^2+\ldots\right)-1}{\left(1+x\ln 4+\frac{x^2}{2!}(\ln 4)^2+\ldots\right)-\left(1+x\ln 3+\frac{x^2}{2!}(\ln 3)^2+\ldots\right)}$$ $$=\lim_{x\to 0}\frac{\frac{\frac{1}{3}}{1!}(-x)+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!}(-x)^2+\ldots}{\left(x\ln 4+\frac{x^2}{2!}(\ln 4)^2+\ldots\right)-\left(x\ln 3+\frac{x^2}{2!}(\ln 3)^2+\ldots\right)}$$ Dividing numerator & denominator by $x$, $$=\lim_{x\to 0}\frac{-\frac{1}{3}+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!}(x)+\ldots}{\left(\ln 4+\frac{x}{2!}(\ln 4)^2+\ldots\right)-\left(\ln 3+\frac{x}{2!}(\ln 3)^2+\ldots\right)}$$

$$=\frac{-\frac{1}{3}+0}{\left(\ln 4+0\right)-\left(\ln 3+0\right)}$$ $$=\color{red}{-\frac{1}{3\ln\left(\frac{4}{3}\right)}}$$

Answer 3

Hint: $$ \lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}= \lim_{x\to0} \frac {\dfrac{(1-x)^{1/3}-1}{x}} {\dfrac{4^x-1}{x}-\dfrac{3^x-1}{x}} $$ Compute each of the three parts and you're done.