# calculating a limit using taylor series

$$\lim_{x\rightarrow0}\frac{\cos(x) \sin(-x)}{\sin^3 x}$$

I tried to use the identity $\cos(x) \sin(x)=\frac{1}{2}\sin(2x)$

and then use the taylor $\lim_{x\rightarrow0}\frac{x+\alpha(x)}{x^3+\beta(x)}$ where $\alpha(x), \beta(x)$ are the remainders which leads me to infinity but it's not the right answer because I did the same with l'hopital rule and got $-\frac{3}{2}$ (I used it three times!)

Can anyone help me how to solve it using taylor series?

Thanks in advance!

## 3 Answers

$$\lim _{x\rightarrow 0} \frac {\cos x \sin x -x }{\sin^3 x} =\lim _{x\rightarrow 0} \frac { \sin 2x -2x }{2\sin^3 x} =\lim_{x\rightarrow 0} \frac { 2x-\frac {8x^3}{3!} +o(x^3) -2x }{ 2\sin^3 x} =\\ =\lim_{x\rightarrow 0} \frac {-\frac {8x^3}{3!} +o(x^3)}{2\sin^3 x} =\lim_{x\rightarrow 0} -\frac {8x^3}{12\sin^3 x} +\lim _{x\rightarrow 0} \frac { o\left(x^3 \right)}{2\sin^3 x} =-\frac 2 3$$ And by L'Hospital's rule $$\lim _{x\rightarrow 0} \frac {\cos x \sin x -x }{\sin^3 x} =\lim_{x\rightarrow 0} \frac {\sin 2x -2x}{2\sin^3 x} \overset {\text{L'Hospital}} = \lim_{x\rightarrow 0} \frac{2\cos 2x -2 }{6\sin^2 x\cos x} =\\ =4\lim_{x\rightarrow 0} \frac {-\sin^2 x}{6\sin^2 x\cos x} =-\frac 2 3$$

• Although my change from $A \overset{L'Hopital} = B$ to $A\overset{\text{L'Lopital}} = B$ required me to add some characters that were not there before, the software summarized my edit by saying it consisted of deletion of 292 characters. The inclusion of so many purposeless extra characters in MathJax code would have amazed me if I hadn't know that some people use software packages to write the code for them and those packages write like a psychotic. But I had thought that was mostly among newbies. And then I see a 16.8k reputation. The code here is fairly routine. – Michael Hardy Jun 25 '17 at 22:02

We have $\sin^{3}x=x^3+\cdots$ and \begin{eqnarray*} \frac{1}{2} \sin(2x)=\frac{1}{2} \left(2x-\frac{(2x)^3}{3!}+\cdots \right) \end{eqnarray*} So \begin{eqnarray*} \lim_{x\rightarrow0}\frac{\cos x\cdot \sin-x}{\sin^{3}x} = \lim_{x\rightarrow0}\frac{-2x^3 /3 +\cdots }{x^3 +\cdots } = \color{red}{\frac{-2}{3}}. \end{eqnarray*}

Using power series, it is pretty simple to show that $$\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16$$ Then \begin{align} \lim_{x\to0}\frac{\cos(x)\sin(x)-x}{\sin^3(x)} &=\lim_{x\to0}\left(\frac{\cos(x)\sin(x)-\sin(x)}{\sin^3(x)}-\frac{x-\sin(x)}{\sin^3(x)}\right)\\ &=\lim_{x\to0}\frac{\cos(x)-1}{\sin^2(x)}-\left(\lim_{x\to0}\frac{x}{\sin(x)}\right)^3\lim_{x\to0}\frac{x-\sin(x)}{x^3}\\ &=\lim_{x\to0}\frac{-1}{1+\cos(x)}-1^3\cdot\frac16\\[3pt] &=-\frac12-\frac16\\[3pt] &=-\frac23 \end{align}