Why limit of $\lim_{x \to 0}(x-\ln(x+1)+\cos(xe^{-x}))^{\frac{9\cot^{3}(x)}{2}}$ is $e^3$? 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!
 A: $$
\ln[\lim_{x\to 0} (x-\ln(x+1)-\cos(xe^{-x})]^ \frac{9\cot^3x}{2} 
\\=\lim_{x\to 0} \ln[(x-\ln(x+1)-\cos(xe^{-x})]^ \frac{9\cot^3x}{2}
\\
=\lim_{x\to 0}\big\{ \big[\frac{9\cot^3x}{2}\big]\big[\ln((x-\ln(x+1)-\cos(xe^{-x}))\big] \big\}
\\
=\lim_{x\to 0}\big\{\frac{9}{2}\big[\frac{1}{x^3}-\frac{1}{x}+\frac{4x}{15}+...\big]\big[\frac{2x^3}{3}-\frac{17x^4}{24}+...\big]\big\}
\\=3
$$
A: 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,
$$\lim_{x\to 0}\frac 92 \frac{\cos^3 x}{\sin^3 x}\ln \left(x-\ln(x+1)+\cos(xe^{-x})\right) = \frac 92 \cdot \frac 23 = 3$$.
