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. 
 A: Hint: 
as $u\to 0,$ $(\ln (1+u))/u \to 1,$ $u(\cot u) \to 1,$ $(\tan u)/u \to 1,$ $(\sin u)/u \to 1.$ No power series necessary to justify these, just well known limits and derivatives.
A: Use equivalents:
$$\ln(1+u)\sim_0 u, \enspace\sqrt{1+u}-1\sim_0 \dfrac u2, \enspace \sin u\sim_0 u, \enspace \cos u\sim_0 1,  \enspace\cot u\sim_0 \dfrac1u, \enspace \tan u\sim_0 u$$ 
hence $$\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}\sim_0 \frac{x^3\cfrac1{x^3}x^4}{\sqrt2\Bigl(\sqrt{1+\frac{x^2}2}-1\Bigr)x^2}\sim_0 \frac{x^2}{\sqrt2\dfrac{x^2}{4}}=2\sqrt2.$$
A: 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 $f \sim g$,
$$ \lim_{x \to 0} \frac{f(x)}{g(x)} = 1
$$
A: This is exactly the kind of question which "looks very difficult" and at the same time is "extremely easy to answer". It appears that it is specially crafted to generate a complicated looking expression in order to intimidate a casual reader/student. The following evaluation shows that the difficulty is only superficial:
\begin{align}
L &= \lim_{x \to 0}\frac{\log(1 + \sin^{3}x \cos^{2}x)\cot(\log^{3}(1 + x))\tan^{4}x}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\log(1 + x^{2})}\tag{1}\\
&= \lim_{x \to 0}\dfrac{\log(1 + \sin^{3}x \cos^{2}x)\cot(\log^{3}(1 + x))\cdot\dfrac{\tan^{4}x}{x^{4}}\cdot x^{4}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot\dfrac{\log(1 + x^{2})}{x^{2}}\cdot x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\log(1 + \sin^{3}x \cos^{2}x)\cot(\log^{3}(1 + x))\cdot 1 \cdot x^{4}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot 1 \cdot x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\log(1 + \sin^{3}x \cos^{2}x)x^{2}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\tan(\log^{3}(1 + x))}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{\log(1 + \sin^{3}x \cos^{2}x)}{\sin^{3}x \cos^{2}x}\cdot \sin^{3}x \cos^{2}x\cdot x^{2}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot\dfrac{\tan(\log^{3}(1 + x))}{\log^{3}(1 + x)}\cdot \dfrac{\log^{3}(1 + x)}{x^{3}}\cdot x^{3}}\notag\\
&= \lim_{x \to 0}\dfrac{1\cdot \sin^{3}x \cos^{2}x\cdot x^{2}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot1\cdot 1\cdot x^{3}}\notag\\
&= \lim_{x \to 0}\dfrac{\sin^{3}x \cdot 1}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})x}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{\sin^{3}x}{x^{3}} \cdot x^{3}}{\dfrac{\sin(\sqrt{x^{2} + 2} -\sqrt{2})}{\sqrt{x^{2} + 2} -\sqrt{2}}\cdot(\sqrt{x^{2} + 2} -\sqrt{2})x}\notag\\
&= \lim_{x \to 0}\dfrac{1 \cdot x^{3}}{1\cdot(\sqrt{x^{2} + 2} -\sqrt{2})x}\notag\\
&= \lim_{x \to 0}\frac{x^{2}}{\sqrt{x^{2} + 2} - \sqrt{2}}\tag{2}\\
&= \lim_{x \to 0}\frac{x^{2}(\sqrt{x^{2} + 2} + \sqrt{2})}{(x^{2} + 2) - 2}\notag\\
&= \lim_{x \to 0}(\sqrt{x^{2} + 2} + \sqrt{2})\notag\\
&= \sqrt{2} + \sqrt{2}\notag\\
&= 2\sqrt{2}\notag
\end{align}
The simplification of the complicated expression in equation $(1)$ to a very simple expression in equation $(2)$ is done via the use of standard limits $$\lim_{x \to 0}\cos x = \lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\tan x}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ In evaluation of limits it is important to figure out the use of standard limits to simplify a complicated expression.
A: $\sin(1+x)=x+\mathcal O(x^3)\;$,  and likewise for the logarithmic function, so:
$$\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}\cong$$
$$\frac{\left(\sin^3x\cos^2x\right)\cot\left(x^3\right)\left(x^4\right)}{\left(\sqrt{x^2+2}-\sqrt2\right)\left(x^2\right)}=\cos x^3\;\frac{x^3}{\sin x^3}\;\cos^2x\;\frac{\sin^3x}{x^3}\left(\sqrt{x^2+2}+\sqrt2\right)\longrightarrow$$
$$\xrightarrow[x\to0]{}1\cdot1\cdot1\cdot1\cdot2\sqrt2=2\sqrt2=\sqrt8$$
