Evaluate $\lim_{x \to 0} \frac{\ln\left[\frac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\right]}{x^3}$ without L'Hôpital/Taylor/differentiation/integration

Solution 1:

You have a limit of type : $$\lim_{x \to 0} \dfrac{\ln f(x)}{g(x)}$$ with : $$\lim_{x \to 0} f(x) = 1$$ Write : $$\lim_{x \to 0} \dfrac{\ln f(x)}{f(x) - 1} \dfrac{f(x) - 1}{g(x)}$$ and you have : $$\lim_{x \to 0} \dfrac{\ln f(x)}{f(x) - 1} = 1$$ The other part : $$\lim_{x \to 0} \dfrac{f(x) - 1}{g(x)}$$ is a simple limit of a fraction.

Solution 2:

Blockquote $\displaystyle\lim_{x \to 0} \frac{\ln\left[\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\right]}{x^{3}}$
$=\displaystyle\lim_{x \to 0}\dfrac{\ln(1-3x)-\ln(1+3x)+3\ln(1+x)-3\ln(1-x)}{x^{3}}$
we use Taylor series
$\displaystyle\lim_{x \to 0}\dfrac{(-3x-\dfrac{(3x)^{2}}{2}-\dfrac{(3x)^{3}}{3}-\dfrac{(3x)^{4}}{4}-\dfrac{(3x)^{5}}{5}-\cdots)-(3x-\dfrac{(3x)^{2}}{2}+\dfrac{(3x)^{3}}{3}-\dfrac{(3x)^{4}}{4}+\dfrac{(3x)^{5}}{5}-\cdots)+3(x-\dfrac{(x)^{2}}{2}+\dfrac{(x)^{3}}{3}-\dfrac{(x)^{4}}{4}+\dfrac{(x)^{5}}{5}-\cdots)-3(-x-\dfrac{(x)^{2}}{2}-\dfrac{(x)^{3}}{3}-\dfrac{(x)^{4}}{4}-\dfrac{(x)^{5}}{5}-\cdots)}{x^{3}}$
$\displaystyle\lim_{x \to 0}\dfrac{-3x-\dfrac{(3x)^{2}}{2}-\dfrac{(3x)^{3}}{3}-\dfrac{(3x)^{4}}{4}-\dfrac{(3x)^{5}}{5}-\cdots-3x+\dfrac{(3x)^{2}}{2}-\dfrac{(3x)^{3}}{3}+\dfrac{(3x)^{4}}{4}-\dfrac{(3x)^{5}}{5}+\cdots+3x-3\dfrac{(x)^{2}}{2}+3\dfrac{x)^{3}}{3}-3\dfrac{(x)^{4}}{4}+3\dfrac{(x)^{5}}{5}-\cdots+3x+3\dfrac{(x)^{2}}{2}+3\dfrac{(x)^{3}}{3}+3\dfrac{(x)^{4}}{4}+3\dfrac{(x)^{5}}{5}+\cdots}{x^{3}}$
$=\displaystyle\lim_{x \to 0}\dfrac{-2(\dfrac{(3x)^{3}}{3}+\dfrac{(3x)^{5}}{5}+\dfrac{(3x)^{7}}{7}+\cdots +6(\dfrac{(x)^{3}}{3}+\dfrac{(x)^{5}}{5}+\dfrac{(x)^{7}}{7}+\cdots}{x^{3}}$
$=\displaystyle\lim_{x \to 0}-2(\dfrac{(3x)^{3}}{3x^{3}}+\dfrac{(3x)^{5}}{5x^{3}}+\dfrac{(3x)^{7}}{7x^{3}}+\cdots +6(\dfrac{(x)^{3}}{3x^{3}}+\dfrac{(x)^{5}}{5x^{3}}+\dfrac{(x)^{7}}{7x^{3}}+\cdots$
$=\displaystyle\lim_{x \to 0}-2(\dfrac{27(x)^{3}}{3x^{3}}+\dfrac{((3)^{5}(x)^{5}}{5x^{3}}+\dfrac{(3)^{7}(x)^{7}}{7x^{3}}+\cdots +6(\dfrac{(x)^{3}}{3x^{3}}+\dfrac{(x)^{5}}{5x^{3}}+\dfrac{(x)^{7}}{7x^{3}}+\cdots$
$=\displaystyle\lim_{x \to 0}-2(9+\dfrac{((3)^{5}(x)^{2}}{5}+\dfrac{(3)^{7}(x)^{4}}{7}+\cdots +6(\dfrac{1}{3}+\dfrac{(x)^{2}}{5}+\dfrac{(x)^{4}}{7}+\cdots$
$=-2(9)+6(\dfrac{1}{3})$
$=-18+2$ $=-16$

Solution 3:

The use of an identity is meaningful for the problem.

$arctanh(z)=\frac{ln(1+z)-ln(1-z)}{2}$

With this Your problem writes as

$lim_{x\rightarrow 0}\frac{6 arctanh(x)-2 arctanh(3 x)}{x^3}$

This works straight forward using

$6 arctanh(x)-2arctanh(3x)=-3 ln(1-x)+3 ln(1+x)-ln(1+3x)-ln(1-3x)=ln(\frac{(1+x)^3(1-3x)}{(1-x)^3(1+3x)})$

using $ln(a b)=ln(a)+ln(b)$ and $ln(\frac{a}{b})=ln(a)-ln(b)$.

Now $arctanh$ has a series representation that is not necessarily a Taylor series. It is a transcendental function. This series representation can be truncated with little error for small values of $x$.

$arctanh(x)=\sum_{k=1}^{\infty}\frac{x^{2k-1}}{(2k-1)!}\approx x- \frac{x^3}{3} + $ error of order $5$

$\frac{6(x-\frac{x^3}{3})-2 (3 x - \frac{(3 x)^3}{3})}{x^3}$

$\frac{(2-18)x^3}{x^3}=-16$

Because of the series representation, everything is like required: no Taylor expansion, no l'Hopital, no differentiation or integration.

Indeed used is an identity for logarithms and the series for Arcustangenshyperbolicus. Because we have two terms for the numerator both cancel in linear order as x gets smaller and the third-order term remains as is suggested already by the given fraction. As x is taken closer to zero the next order is five with coefficient $-96$ but with denominator, this reduces to second order. This cancels down to the constant term and that is $-16$ as shown in my calculation. A little disambiguation remains since the Taylor series around 0 coincides with the series representation of Arcustangenshyperbolicus. For limits usually, only the Taylor expansion up to noncritical orders is used by definition. Mind this is per se only valid for $x>0$. For $x<0$ a similar calculation is possible and the results are the same.