<s>[INST] How to find:$\lim_{x\to 0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$ 
Evalute: $$\lim_{x\to
 0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$$

My attempt:
I used the standard limits from the table:$$\lim_{x\to 0}\frac{\sin x}{x}=1,\;\;\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2},\;\;\lim_{x\to 0}\frac{\tan x}{x}=1,\;\;\lim_{x\to 0}\frac{e^x-1}{x}=1$$
$$$$
$L=\displaystyle\lim_{x\to 0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}=$
$$$$
$\displaystyle\lim_{x\to 0}\left[\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{e^{1-\cos^3x}-e^{1-\cos^4x}}\cdot\left(\frac{e^{1-\cos^3x}-1}{1-\cos^3x}\cdot\frac{1-\cos^3x}{x^2}-\frac{e^{1-\cos^4x}-1}{1-\cos^4x}\cdot\frac{1-\cos^4x}{x^2}\right)\cdot\frac{x}{\arctan x}\right]$
Substitution: $$[t=\arctan x\implies x=\tan t\;\;\&\;\; x\to 0\implies t\to 0]$$
$$\lim_{x\to 0}\frac{x}{\arctan x}\iff\lim_{t\to 0}\frac{\tan t}{t}=1$$
The next step:$$1-\cos^3x=(1-\cos x)(1+\cos x+\cos^2x)$$
$$1-\cos^4x=(1-\cos x)(1+\cos x)(1+\cos^2x)$$
Now I obtained:
$$L=1\cdot\left(1\cdot\frac{3}{2}-1\cdot 2\right)\cdot 1=-\frac{1}{2}$$
Is this correct?
 [/INST] Here is another way after your first step (removing the $\sin() $). Note that if $A, B$ tend to $1$ then $$A-B=B\cdot\frac{\exp(\log A - \log B) - 1}{\log A - \log B} \cdot(\log A - \log B) $$ and the first two factors above tend to $1$. Therefore $A-B$ can be replaced by $\log A - \log B$.
Thus the numerator in your case gets replaced by $$(1-\cos^3x)-(1-\cos^4x)=-\cos^3x(1-\cos x) $$ and therefore we can safely replace it by $-(1-\cos x) $. The denominator $x\arctan x$ can be replaced by $x^2$ because of limit $\lim_{x\to 0}\dfrac {\arctan x} {x} =1$. It should be now clear that the desired limit is $-1/2$.

See the usage of this technique here, here and here. 
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