Why limit of $\lim_{x \to 0}(x-\ln(x+1)+\cos(xe^{-x}))^{\frac{9\cot^{3}(x)}{2}}$ is $e^3$?

Question

I'm trying to find: $$\lim_{x \to 0}(x-\ln(x+1)+\cos(xe^{-x}))^{\frac{9cot^{3}(x)}{2}}$$

Considering: $$\lim_{x \to 0}(x-\ln(x+1)+\cos(xe^{-x})) = 1$$

I thought that was the answer but WolframAlpha gives me $e^3$. Now I realize that I should make a Taylor series to resolve indeterminate form of $\cot$ which is: $$\cot(x) = \left[\frac{1}{0}\right] = \frac{1}{x} - \left(\frac{x}{3} + \frac{x^{3}}{45} + ... \right)$$ but I still don't understand how do we get $e^3$ answer!

Answer 1

The "surprise" is caused by the term $xe^{-x}$ within the cosine.

First you may take the logarithm and consider

$$\frac 92 \frac{\cos^3 x}{\sin^3 x}\ln \left(x-\ln(x+1)+\cos(xe^{-x})\right)\stackrel{x\to 0}{\sim}\frac 92\frac{\ln \left(x-\ln(x+1)+\cos(xe^{-x})\right)}{x^3}$$

Now using $\lim_{y\to 0}\frac{\log(\color{blue}{1}+y)}{y} = 1$ consider only

$$\frac{x-\ln(x+1)+\cos(xe^{-x})\color{blue}{-1}}{x^3}$$

Now, you may apply Taylor and note that $o(x^3e^{-3x}) = o(x^3)$:

$$\frac{x-\left(x-\frac{x^2}{2}+ \frac{x^3}{3}+o(x^3)\right)+1-\frac{x^2e^{-2x}}{2}+o(x^3e^{-3x})-1}{x^3}$$ $$= \frac 12\frac{1-e^{-2x}}{x} - \frac 13 + o(1)\stackrel{x\to 0}{\rightarrow}1-\frac 13 = \frac 23$$

Hence,

$$\lim_{x\to 0}\frac 92 \frac{\cos^3 x}{\sin^3 x}\ln \left(x-\ln(x+1)+\cos(xe^{-x})\right) = \frac 92 \cdot \frac 23 = 3$$.

Answer 2

$$ \ln[\lim_{x\to 0} (x-\ln(x+1)-\cos(xe^{-x})]^ \frac{9\cot^3x}{2} \\=\lim_{x\to 0} \ln[(x-\ln(x+1)-\cos(xe^{-x})]^ \frac{9\cot^3x}{2} \\ =\lim_{x\to 0}\big\{ \big[\frac{9\cot^3x}{2}\big]\big[\ln((x-\ln(x+1)-\cos(xe^{-x}))\big] \big\} \\ =\lim_{x\to 0}\big\{\frac{9}{2}\big[\frac{1}{x^3}-\frac{1}{x}+\frac{4x}{15}+...\big]\big[\frac{2x^3}{3}-\frac{17x^4}{24}+...\big]\big\} \\=3 $$

Answer 3

When you compose Taylor series, build one piece at the time and you can obtain more than the limit itself (which can be very interesting. Working your case $$y=\big[x-\ln(x+1)+\cos(xe^{-x})\big]^{\frac{9\cot^{3}(x)}{2}}$$ $$\log(y)={\frac{9}{2}}\cot^{3}(x)\log\big[x-\ln(x+1)+\cos(xe^{-x})\big]$$ $$\log(x+1)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+O\left(x^7\right)$$ $$e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\frac{x^5}{120}+\frac{x^6}{720}+O\left(x^7\right)$$ $$xe^{-x}=x-x^2+\frac{x^3}{2}-\frac{x^4}{6}+\frac{x^5}{24}-\frac{x^6}{120}+\frac{x^7}{720}+O\left(x^8\right)$$ $$\cos(xe^{-x})=1-\frac{x^2}{2}+x^3-\frac{23 x^4}{24}+\frac{x^5}{2}-\frac{x^6}{720}-\frac{109 x^7}{360}+\frac{15121 x^8}{40320}+O\left(x^9\right)$$ $$x-\ln(x+1)+\cos(xe^{-x})=1+\frac{2 x^3}{3}-\frac{17 x^4}{24}+\frac{3 x^5}{10}+\frac{119 x^6}{720}+O\left(x^7\right)$$ $$\log\big[x-\ln(x+1)+\cos(xe^{-x})\big]=\frac{2 x^3}{3}-\frac{17 x^4}{24}+\frac{3 x^5}{10}-\frac{41 x^6}{720}+O\left(x^7\right)$$ $$\cot(x)=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2 x^5}{945}+O\left(x^7\right)$$ $$\cot^3(x)=\frac{1}{x^3}-\frac{1}{x}+\frac{4 x}{15}+\frac{x^3}{945}+O\left(x^5\right)$$ $$\log(y)=3-\frac{51 x}{16}-\frac{33 x^2}{20}+\frac{469 x^3}{160}+O\left(x^4\right)$$ $$y=e^{\log(y)}=e^3\left(1-\frac{51 x}{16}+\frac{8781 x^2}{2560}+\frac{114403 x^3}{40960}+O\left(x^4\right) \right)$$ Just for the fun, put on the same graph $y$ and the above truncated expression. You will see that the curves overlap for $0 \leq x \leq 0.2$.

This means that if you have to approximatively solve for $x$ the equation $y=15$, you just need to solve a cubic equation in $x$; this would give $x=0.08846$ while the exact solution would be $0.08819$.