# Limit with Taylor formula

• Nov 9th 2008, 10:59 AM
xkyve
Limit 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 AM
Jhevon
Quote:

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

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

hint:

$\displaystyle \sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - + \cdots$ for all $\displaystyle x$

and

$\displaystyle \tan x = x + \frac {x^3}{3} + \frac {2x^5}{15} + \cdots$ for $\displaystyle |x| < \frac {\pi}2$
• Nov 9th 2008, 11:40 AM
xkyve
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 AM
Mathstud28
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 AM
xkyve
thank you :)
one more question, did u do those computations by hand?
• Nov 10th 2008, 03:11 AM
Mathstud28
Quote:

Originally Posted by xkyve
thank you :)
one more question, did u do those computations by hand?

Up to the third derivative. And then I was like screw this and used my calculator.
• Nov 10th 2008, 04:06 AM
xkyve
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 PM
Mathstud28
Quote:

Originally Posted by xkyve
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...

There are other methods, but none that are using the Taylor method. Because that is the Taylor method.
• Nov 11th 2008, 02:58 AM
xkyve
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 AM
Mathstud28
Quote:

Originally Posted by xkyve
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

Mathcad, TI-89, or Mathematica are all good choices.