# 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.

• Well done on using mathjax, just a tip: put slashes before your \sin and \cos to make them display a little nicer: $\sin$ $\cos$. – nathan.j.mcdougall Apr 16 '15 at 19:16
• You need to go to order $2$ in $(1+x)^{1/3}$ – egreg Apr 16 '15 at 19:19
• @Essam did you even read the post? It says very clearly "limits without lhopital" and using with Taylor series. – user223391 Apr 16 '15 at 19:19
• Or you could try multiplying by the denominator's conjugate, giving $\sin^2(x)$, which is much easier to work with. – nathan.j.mcdougall Apr 16 '15 at 19:20

Hint : develop to order 2. If you have 0/0, you did not take enough terms into account to understand how the numerator/denominator converge (in fact, the significant terms are still hidden in the $o(x)$)
$$\frac{(1+3x)^{1/3}-\sin x-1}{1-\cos x}$$
$$=\frac{1+\frac13\cdot3x+\frac13\left(\frac13-1\right)\dfrac{(3x)^2}{2!}+O(x^3)-\left[x+O(x^3)\right]-1}{1-\left[1-\dfrac{x^2}2+O(x^4)\right]}$$
$$=\frac{-x^2+O(x^3)}{\dfrac{x^2}2+O(x^4)}$$