<s>[INST] Find $\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$ $$\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$$
I don't think L'hospital's rule will make the problem easy. (I am afraid to differentiate the numerator). The given limit has a $\frac{0}{0}$ form. I tried using taylor series but the it made the problem more complicated. 
 [/INST] Using infinitesimals of equal order,
$$\begin{align*} \lim_{x\to 0}\frac{\log(1+\sin^3x \cos^2x)\cot(\log(1+x)^3)\tan^4 x}{\sin(\sqrt{x^2+2}-\sqrt{2})\log(1+x^2)}
&= \lim_{x\to 0}\frac{\log(1+\sin^3x \cos^2x)\cot(\log(1+x)^3)x^4}{\sin(\sqrt{x^2+2}-\sqrt{2})\log(1+x^2)} \\
&= \lim_{x\to 0}\frac{ \sin^3x \cos^2x \cot(\log(1+x)^3)x^4}{(\sqrt{x^2+2}-\sqrt{2})x^2} \\
&= \lim_{x\to 0}\frac{ x^3 \cot(\log(1+x)^3)x^4}{(\sqrt{x^2+2}-\sqrt{2})x^2} \\
&= \lim_{x\to 0}\frac{x^3 x^{-3} x^4}{(\sqrt{x^2+2}-\sqrt{2})x^2}
\end{align*}$$
where $\cot(\log(1+x)^3) \sim x^{-3}$ since $\cot \sim x^{-1}, \log(1+x)^3 \sim x^3$. Some of the other infinitesimals I've used:
$$\sin x \sim x, \cos x \sim 1, \log(1+x) \sim x, \tan x \sim x
$$
It is easily checked with l'Hopital that if $f \sim g$,
$$ \lim_{x \to 0} \frac{f(x)}{g(x)} = 1
$$
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