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
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:
$\displaystyle \lim_{x\to 0}\frac{sinh(x^{3})-sin(x^{3})}{x^{9}}$
$\displaystyle \text{senh}(x)$?
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
$\displaystyle \begin{aligned}\lim_{x\to{0}}\frac{\sinh\left(x^3\ right)-\sin\left(x^3\right)}{x^9}&=\lim_{x\to{0}}\frac{\l eft(x^3+\frac{x^9}{6}+\cdots\right)-\left(x-\frac{x^9}{6}+\cdots\right)}{x^9}\\
&=\lim_{x\to{0}}\frac{\frac{x^9}{3}+\cdots}{x^9 }\\
&=\frac{1}{3}\end{aligned}$
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?