"Calculate the limit using Taylor's formula". I know that the limit should be 1/30. Problem is how to prove it.

$\displaystyle

\lim_{x\rightarrow0}\frac{tan(sinx) - sin(tanx)}{x^7}

$

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- Nov 9th 2008, 10:59 AMxkyveLimit with Taylor formula
"Calculate the limit using Taylor's formula". I know that the limit should be 1/30. Problem is how to prove it.

$\displaystyle

\lim_{x\rightarrow0}\frac{tan(sinx) - sin(tanx)}{x^7}

$ - Nov 9th 2008, 11:10 AMJhevon
- Nov 9th 2008, 11:40 AMxkyve
i know the expansions and i found the value 17/315 + 1/7! for the limit but it's incorrect. i think i must find a path to write those trig. functions in another way.

- Nov 9th 2008, 11:58 AMMathstud28
Let $\displaystyle f(x)=\tan(\sin(x))$ and $\displaystyle g(x)=\sin(\tan(x))$. Then

$\displaystyle f(0)=0$ ----$\displaystyle g(0)=0$

$\displaystyle f'(0)=1$ ----$\displaystyle g'(0)=1$

$\displaystyle f''(0)=0$ ----$\displaystyle g''(0)=0$

$\displaystyle f^{(3)}(0)=1$ ----$\displaystyle g^{(3)}(0)=1$

$\displaystyle f^{(4)}(0)=0$---- $\displaystyle g^{(4)}(0)=0$

$\displaystyle f^{(5)}(0)=-3$ ----$\displaystyle g^{(5)}(0)=-3$

$\displaystyle f^{(6)}(0)=0$ ----$\displaystyle g^{(6)}(0)=0$

$\displaystyle f^{(7)}(0)=-107$---- $\displaystyle g^{(7)}(0)=-275$

$\displaystyle \therefore\quad\tan(\sin(x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107x^7}{5040}\pm\cdots$

and $\displaystyle \sin(\tan(x))=x+\frac{x^3}{6}-\frac{x^5}{4}-\frac{275x^7}{5040}\pm\cdots$

$\displaystyle \begin{aligned}\therefore\quad&\lim_{x\to{0}}\frac {\tan(\sin(x))-\sin(\tan(x))}{x^7}\\

&=\lim_{x\to{0}}\frac{x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107x^7}{5040}\pm\cdots-\left(x+\frac{x^3}{6}-\frac{x^5}{4}-\frac{275x^7}{5040}\pm\cdots\right)}{x^7}\\

&=\lim_{x\to{0}}\frac{\frac{x^7}{30}\pm\cdots}{x^7 }\\

&=\frac{1}{30}

\end{aligned}$ - Nov 10th 2008, 01:37 AMxkyve
thank you :)

one more question, did u do those computations by hand? - Nov 10th 2008, 03:11 AMMathstud28
- Nov 10th 2008, 04:06 AMxkyve
yeah, those computations are long... anyways, i was wondering if there was another method, one that envolves only a pencil and some paper. :D i'll have to think about it...

- Nov 10th 2008, 12:15 PMMathstud28
- Nov 11th 2008, 02:58 AMxkyve
probably you are right, but i will still try to solve it using taylor but less computations.

doesn't matter, can u recommend me some good software that is able to compute derivatives, limits, and so on?

thanks - Nov 11th 2008, 04:03 AMMathstud28