# HELP! solving limit with taylor and young

• Nov 29th 2008, 03:26 PM
clar
HELP! solving limit with taylor and young
(Crying) i really need someone to solve this limit for me, cause im not sure if i did it right, it has to be done using taylor and young

lim senhx^3 - senx^3 /x^9
x-> 0
• Nov 29th 2008, 03:32 PM
galactus
I must plead ignorance. What is the Taylor and Young method of limits?. I even googled it and found nothing. Do you mean using Taylor series?.
Also, what is sen and senh?. I assume perhaps sin and sinh?.

Is this it:

$\displaystyle \lim_{x\to 0}\frac{sinh(x^{3})-sin(x^{3})}{x^{9}}$
• Nov 29th 2008, 03:37 PM
Mathstud28
Quote:

Originally Posted by clar
(Crying) i really need someone to solve this limit for me, cause im not sure if i did it right, it has to be done using taylor and young

lim senhx^3 - senx^3 /x^9
x-> 0

$\displaystyle \text{senh}(x)$?

Using the fact that i is above the middle finger as is e I assume your question is

$\displaystyle \lim_{x\to{0}}\frac{\sinh\left(x^3\right)-\sin\left(x^3\right)}{x^9}$

Note that $\displaystyle \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$ we can see two things

$\displaystyle \sin\left(x^3\right)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{6n+3}}{(2n+1)!}$

And

\displaystyle \begin{aligned}\sinh\left(x^3\right)&=\sin\left(ix ^3\right)\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n\left(ix\right)^{2n+1}}{(2n+1)!}\\ &=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}\end{ aligned}

So your limit is equivalent to

\displaystyle \begin{aligned}\lim_{x\to{0}}\frac{\sinh\left(x^3\ right)-\sin\left(x^3\right)}{x^9}&=\lim_{x\to{0}}\frac{\l eft(x^3+\frac{x^9}{6}+\cdots\right)-\left(x-\frac{x^9}{6}+\cdots\right)}{x^9}\\ &=\lim_{x\to{0}}\frac{\frac{x^9}{3}+\cdots}{x^9 }\\ &=\frac{1}{3}\end{aligned}
• Nov 29th 2008, 03:38 PM
clar
yes! thats right, sorry its just that im from costa rica, we speak spanish so its diferent
• Nov 29th 2008, 03:41 PM
galactus
So, in Coats Rica Taylor series are called Taylor and Young series?. That's a new one I will have to remember.
• Nov 29th 2008, 03:50 PM
clar
Not exactly, were supose to solve the limit using the usin taylor series and (resto de Young) which i dont know exactly how to translate into english, it would be like the rest of young which sounds dumb. But its basically the error o(x^n). Sorray i dont really know how to explain.

But basically the problem i have with the limit is the answer i dont understand why its one third, what happened to the factorial?
• Nov 29th 2008, 03:51 PM
Mathstud28
Quote:

Originally Posted by clar
Not exactly, were supose to solve the limit using the usin taylor series and (resto de Young) which i dont know exactly how to translate into english, it would be like the rest of young which sounds dumb. But its basically the error o(x^n). Sorray i dont really know how to explain.

But basically the problem i have with the limit is the answer i dont understand why its one third, what happened to the factorial?

$\displaystyle \frac{1}{3!}+\frac{1}{3!}=\frac{1}{6}+\frac{1}{6}= \frac{1}{3}$
• Nov 29th 2008, 03:53 PM
clar
jaja oops (Giggle), i feel really dumb right now, thank u so much
• Nov 29th 2008, 03:54 PM
galactus
Oh, I see. It translates as Big O. The terms which are more or less extraneous. i.e. Not needed because they become small.
• Nov 29th 2008, 03:56 PM
clar
ah big O, ill remember that, thanks a lot