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?
 A: $$\dfrac{x^3 - \sin^3(x)}{x - \ln(1+x) - 1 + \cos(x)} = \dfrac{x - \sin(x)}{x} \times \dfrac{x^2 + x \sin(x) + \sin^2(x)}{1 - \dfrac{\ln(1+x)}{x}- \dfrac{1-\cos(x)}{x}}$$
$$1 - \dfrac{\ln(1+x)}{x}- \dfrac{1-\cos(x)}{x} = 1 - \left(1 - \dfrac{x}2 +\dfrac{x^2}3 - \cdots \right) - \left( \dfrac{\dfrac{x^2}{2!} - \dfrac{x^4}{4!} + \cdots}{x} \right) \\ = \left(\dfrac{x}2 - \dfrac{x^2}3 + \cdots \right) - \left(\dfrac{x}2 - \dfrac{x^3}{4!} + \cdots \right) = -\dfrac{x^2}3 + \mathcal{O}(x^4)$$
$$\dfrac{x - \sin(x)}{x} = \dfrac{x^2}{3!} + \mathcal{O}(x^4)$$
Hence,$$\dfrac{x^3 - \sin^3(x)}{x - \ln(1+x) - 1 + \cos(x)} = \left(\dfrac{x^2}{3!} + \mathcal{O}(x^4) \right) \times \dfrac{x^2 + x \sin(x) + \sin^2(x)}{-\dfrac{x^2}3 + \mathcal{O}(x^4)}$$
Now you should be able to finish it off.
A: 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*}
A: It suffices to transform the original expression as a product of other
expressions of known limits or maybe computed easily. 
\begin{equation*}
\lim_{x\rightarrow 0}\frac{x^{3}-\sin ^{3}x}{x-\ln (1+x)-1+\cos x}%
=\lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}\right) \frac{\left(
x^{2}+x\sin x+\sin ^{2}x\right) }{\left( \frac{\cos x-1+\frac{1}{2}x^{2}}{%
x^{3}}\right) -\left( \frac{\ln (1+x)-x+\frac{1}{2}x^{2}}{x3}\right) }%
=\left( \frac{1}{6}\right) \frac{\left( 0^{2}+0+\left( 0\right) ^{2}\right) 
}{\left( 0\right) -\left( \frac{1}{3}\right) }=0.
\end{equation*}
