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How to find this limit without using L'Hospital rule

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

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closed as off-topic by user147263, user91500, SchrodingersCat, Claude Leibovici, user223391 Nov 29 '15 at 5:21

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  • $\begingroup$ What have you tried? Where exactly do you need help? Any ideas at all? Please use the math notation environment that is provided. It is difficult to see, what limit you actually want to calculate. $\endgroup$ – Alexander Nov 26 '15 at 19:37
  • $\begingroup$ Do you mean: $\lim_{x\to 0}\frac{\frac{1-x}{3}-1}{4^x-3^x}$? $\endgroup$ – Jan Nov 26 '15 at 19:37
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    $\begingroup$ $\lim_{x\to 0}\frac{(1-x)^\frac{1}{3}-1}{4^x-3^x}$, I think. $\endgroup$ – mzp Nov 26 '15 at 19:41
  • $\begingroup$ the solution is given by $$\left( -6\,\ln \left( 2 \right) +3\,\ln \left( 3 \right) \right) ^ {-1} $$ $\endgroup$ – Dr. Sonnhard Graubner Nov 26 '15 at 19:48
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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.

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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 $x$, $$=\lim_{x\to 0}\frac{-\frac{1}{3}+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!}(x)+\ldots}{\left(\ln 4+\frac{x}{2!}(\ln 4)^2+\ldots\right)-\left(\ln 3+\frac{x}{2!}(\ln 3)^2+\ldots\right)}$$

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

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

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