# Thread: HELP! solving limit with taylor and young

1. ## HELP! solving limit with taylor and young

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

2. 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:

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

3. Originally Posted by clar
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}

4. yes! thats right, sorry its just that im from costa rica, we speak spanish so its diferent

5. So, in Coats Rica Taylor series are called Taylor and Young series?. That's a new one I will have to remember.

6. 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?

7. 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?
$\displaystyle \frac{1}{3!}+\frac{1}{3!}=\frac{1}{6}+\frac{1}{6}= \frac{1}{3}$

8. jaja oops , i feel really dumb right now, thank u so much

9. Oh, I see. It translates as Big O. The terms which are more or less extraneous. i.e. Not needed because they become small.

10. ah big O, ill remember that, thanks a lot