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

Question

The question is:

Evaluate the following limit without L'Hôpital, Taylor, differentiation (or integration): $$\lim_{x \to 0} \frac{\ln\left[\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\right]}{x^3}$$

This came up while I was trying to answer to this question. My own answere is there too.

I managed to cancel out most of the terms without L'Hôpital/Taylor/differentiation until I reached the above limit. Then I used a differentiation (not L'Hôpital) and finished with the definition of a derivative. The OP to that question claimed that even differentiation is not allowed. He believes it could be solvable only using such methods that can be derived (algebraically) from something like $\lim_{x\to0} \frac{\sin x}{x}$ or $\lim_{n\to\infty} \left(1+\frac1n\right)^n$, etc. I think that a sandiwch theorem will be fine too.

Using Taylor seriese or L'Hôpital's theorem, the above limit is trivial to solve. And the below is how I solved it using differentiation (which was not allowed either). I am wondering if there's a way to evaluate it even without differentiation, so I am posting here. Thanks.

My method: \begin{align} \lim_{x \to 0} \frac{\ln\left[\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\right]}{x^3} &=\lim_{x \to 0} \frac{\ln\left((1-3x)(1+x)^3\right)-\ln\left((1+3x)(1-x)^3\right)}{x^3}\\ &=\lim_{x \to 0} \frac{\ln(1-6x^2-8x^3-3x^4)-\ln(1-6x^2+8x^3-3x^4)}{x^3}\\ &=\lim_{x \to 0} \frac{\ln(1-6x^2-8x^3)-\ln(1-6x^2+8x^3)}{x^3}\\ &=\lim_{h \to 0} \frac{\ln(1-6h^{2/3}-8h)-\ln(1-6h^{2/3}+8h)}{h}\\ &=\left.\left\{\ln(1-6x^{2/3}-8x)-\ln(1-6x^{2/3}+8x)\right\}'\right|_{x=0}\\ &=\left.\left(\frac{-\frac4{\sqrt[3]x}-8}{1-6x^{2/3}-8x}-\frac{-\frac4{\sqrt[3]x}+8}{1-6x^{2/3}+8x}\right)\right|_{x=0}\\ &=\left.\frac{-16+32x^{2/3}}{(1-6x^{2/3}-8x)(1-6x^{2/3}+8x)}\right|_{x=0}\\ &=-16\\ \end{align}

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Here's the alternative answer that I think may be acceptable. It only uses: $$\frac1{1+x}=1-x+x^2-x^3+\cdots$$ and $$\lim_{x\to0}\frac{\ln(1+x)}x=1$$ that comes from $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$$ So the method-2 is: \begin{align} &\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\\ =&\dfrac{(1-3x)(1+3x+3x^2+x^3)}{(1+3x)\left(1-x(3+3x-x^2)\right)}\\ =&(1-3x)(1+3x+3x^2+x^3)\\ \times&(1-3x+9x^2-27x^3+\cdots)\\ \times&\left(1+x(3+3x-x^2)+x^2(3+3x-x^2)^2+x^3(3+3x-x^2)^3+\cdots\right)\\ =&(1-3x)(1+3x+3x^2+x^3)(1-3x+9x^2-27x^3+\cdots)(1+3x+6x^2+10x^3+\cdots)\\ =&1-16x^3+O(x^4) \end{align} Therefore, \begin{align} \lim_{x \to 0} \frac{\ln\left[\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\right]}{x^3}&=\lim_{x \to 0} \frac{\ln\left(1-16x^3+O(x^4)\right)}{x^3}\\ &=-16\lim_{x \to 0} \frac{\ln\left(1-16x^3\right)}{-16x^3}\\ &=-16\lim_{x\to0}\frac{\ln(1+x)}x\\ &=-16 \end{align}

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Using the Essaidi's answer below, here's the solution that will (I believe) satisfy the requirement of "using only knowledge learned before Calculus" (this is the updated compilation of the requirement to the original problem that I could understand from the OOP's comment to other person below the original question).

\begin{align} &\lim_{x \to 0} \frac{\ln\left[\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\right]}{x^3}\\ &=\lim_{x \to 0} \frac{\ln\left[1+\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}-1\right]}{\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}-1}\frac{\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}-1}{x^3}\\ &=\lim_{y \to 0}\frac{\ln(1+y)}{y}\cdot\lim_{x \to 0}\frac{\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}-1}{x^3}\\ &=1\cdot\lim_{x \to 0}\frac{(1-3x)(1+x)^3-(1+3x)(1-x)^3}{x^3(1+3x)(1-x)^3}\\ &=\lim_{x \to 0}\frac{-16x^3}{x^3}\\ &=-16 \end{align}

Answer 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.

Answer 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
$\displaystyle\lim_{x \to 0}\dfrac{(-3x-\dfrac{(3x)^{2}}{2}-\dfrac{(3x)^{3}}{3}-\dfrac{(3x)^{4}}{4}-\dfrac{(3x)^{5}}{5}-\cdots)-(3x-\dfrac{(3x)^{2}}{2}+\dfrac{(3x)^{3}}{3}-\dfrac{(3x)^{4}}{4}+\dfrac{(3x)^{5}}{5}-\cdots)+3(x-\dfrac{(x)^{2}}{2}+\dfrac{(x)^{3}}{3}-\dfrac{(x)^{4}}{4}+\dfrac{(x)^{5}}{5}-\cdots)-3(-x-\dfrac{(x)^{2}}{2}-\dfrac{(x)^{3}}{3}-\dfrac{(x)^{4}}{4}-\dfrac{(x)^{5}}{5}-\cdots)}{x^{3}}$
$\displaystyle\lim_{x \to 0}\dfrac{-3x-\dfrac{(3x)^{2}}{2}-\dfrac{(3x)^{3}}{3}-\dfrac{(3x)^{4}}{4}-\dfrac{(3x)^{5}}{5}-\cdots-3x+\dfrac{(3x)^{2}}{2}-\dfrac{(3x)^{3}}{3}+\dfrac{(3x)^{4}}{4}-\dfrac{(3x)^{5}}{5}+\cdots+3x-3\dfrac{(x)^{2}}{2}+3\dfrac{x)^{3}}{3}-3\dfrac{(x)^{4}}{4}+3\dfrac{(x)^{5}}{5}-\cdots+3x+3\dfrac{(x)^{2}}{2}+3\dfrac{(x)^{3}}{3}+3\dfrac{(x)^{4}}{4}+3\dfrac{(x)^{5}}{5}+\cdots}{x^{3}}$
$=\displaystyle\lim_{x \to 0}\dfrac{-2(\dfrac{(3x)^{3}}{3}+\dfrac{(3x)^{5}}{5}+\dfrac{(3x)^{7}}{7}+\cdots +6(\dfrac{(x)^{3}}{3}+\dfrac{(x)^{5}}{5}+\dfrac{(x)^{7}}{7}+\cdots}{x^{3}}$
$=\displaystyle\lim_{x \to 0}-2(\dfrac{(3x)^{3}}{3x^{3}}+\dfrac{(3x)^{5}}{5x^{3}}+\dfrac{(3x)^{7}}{7x^{3}}+\cdots +6(\dfrac{(x)^{3}}{3x^{3}}+\dfrac{(x)^{5}}{5x^{3}}+\dfrac{(x)^{7}}{7x^{3}}+\cdots$
$=\displaystyle\lim_{x \to 0}-2(\dfrac{27(x)^{3}}{3x^{3}}+\dfrac{((3)^{5}(x)^{5}}{5x^{3}}+\dfrac{(3)^{7}(x)^{7}}{7x^{3}}+\cdots +6(\dfrac{(x)^{3}}{3x^{3}}+\dfrac{(x)^{5}}{5x^{3}}+\dfrac{(x)^{7}}{7x^{3}}+\cdots$
$=\displaystyle\lim_{x \to 0}-2(9+\dfrac{((3)^{5}(x)^{2}}{5}+\dfrac{(3)^{7}(x)^{4}}{7}+\cdots +6(\dfrac{1}{3}+\dfrac{(x)^{2}}{5}+\dfrac{(x)^{4}}{7}+\cdots$
$=-2(9)+6(\dfrac{1}{3})$
$=-18+2$ $=-16$

Answer 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.