<s>[INST] Elementary Question about limits Prove 
$$
\lim_{x \to 0}\frac{x^3 - \sin^3x}{x - \ln{(1+x)} - 1 + \cos x} = 0
$$
Obviously, Using l'Hôpitals rule, we can evaluate this limit. But, taking derivatives of such functions is such a mess. Anyone sees a trick to do this problem faster? any ideas?
 [/INST] We have $x^3-\sin^3x=(x-\sin x)(x^2+x\sin x+\sin^2x)\approx\frac{x^3}{6}(3x^2)=\displaystyle\frac{x^5}{2}$ as $x\to 0$.
Hence
\begin{equation*}
\begin{array}{lll}
\displaystyle\lim_{x \to 0}\frac{x^3 - \sin^3x}{x - \ln{(1+x)} - 1 + \cos x} &=&\displaystyle\lim_{x\to 0}\frac{x^5}{2(x - \ln{(1+x)} - 1 + \cos x)}\\
&\overset{H}{=}&\displaystyle\lim_{x \to 0}\frac{5x^4}{2(1-\frac{1}{1+x}-\sin x)}\\
&\overset{H}{=}&\displaystyle\lim_{x \to 0}\frac{10x^3}{\frac{1}{(1+x)^2}-\cos x}\\
&\overset{H}{=}&\displaystyle\lim_{x \to 0}\frac{30x^2}{\frac{-2}{(1+x)^3}+\sin x}\\
&=&\displaystyle\frac{0}{\frac{-2}{(1+0)^3}+0}=0.
\end{array}
\end{equation*}
</s>