Find $\lim\limits_{x \to 0} \frac{(1+3x)^{1/3}-\sin(x)-1}{1-\cos(x)}$ I would like to find using Taylor series :
$$\lim\limits_{x \to 0} \frac{(1+3x)^{1/3}-\sin(x)-1}{1-\cos(x)}$$
So I compute the taylor series of the terms at the order $1$ :
$(1+3x)^{1/3}=1+x+o(x)$ and $-\sin x -1=-1-x+o(x)$ and $1-\cos(x)$ does not have a taylor series at the order $1$ so we have $0$ at the numerator and denominator when we search the limit for $x=0$, according to wolfram we should find $-2$, how is it possible ?
Thank you.

I will show another way to compute this limit. Let us transform the original expression as follows:
\begin{eqnarray*}
\frac{(1+3x)^{\frac{1}{3}}-1-(\sin x)}{1-(\cos x)} &=&\frac{\left( (1+3x)^{%
\frac{1}{3}}-1-x\right) -\left( \sin x-x\right) }{1-(\cos x)} \\
&=&\frac{9\times \left( \frac{(1+3x)^{\frac{1}{3}}-1-\frac{1}{3}(3x)}{%
(3x)^{2}}\right) -\left( \frac{\sin x-x}{x^{2}}\right) }{\left( \frac{1-\cos
x}{x^{2}}\right) }.
\end{eqnarray*}
Using standard limits
\begin{equation*}
\lim_{u\rightarrow 0}\frac{(1+u)^{\frac{1}{3}}-1-\frac{1}{3}u}{u^{2}}=-\frac{%
1}{9},\ \ \ and\ \ \ \ \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{2}}=0,\ \ \ \
and\ \ \ \ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2},
\end{equation*}
the required limit follows
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{(1+3x)^{\frac{1}{3}}-1-(\sin x)}{1-(\cos x)} &=&%
\frac{9\times \lim\limits_{x\rightarrow 0}\left( \frac{(1+3x)^{\frac{1}{3}%
}-1-\frac{1}{3}(3x)}{(3x)^{2}}\right) -\lim\limits_{x\rightarrow 0}\left( 
\frac{\sin x-x}{x^{2}}\right) }{\lim\limits_{x\rightarrow 0}\left( \frac{%
1-\cos x}{x^{2}}\right) } \\
&=&\frac{9\times \left( -\frac{1}{9}\right) -\left( 0\right) }{\left( \frac{1%
}{2}\right) }=-2.
\end{eqnarray*}
The standard limits can be computed using Taylor series, or l'Hospital's rule (or none of them!).
