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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.

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

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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!).

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$$\frac{(1+3x)^{1/3}-\sin x-1}{1-\cos x}$$

$$=\frac{1+\frac13\cdot3x+\frac13\left(\frac13-1\right)\dfrac{(3x)^2}{2!}+O(x^3)-\left[x+O(x^3)\right]-1}{1-\left[1-\dfrac{x^2}2+O(x^4)\right]}$$

$$=\frac{-x^2+O(x^3)}{\dfrac{x^2}2+O(x^4)}$$

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