1. ## limit problem 4_34

Help me to solve this problem

Use Taylor's theorem to compute :

$\displaystyle lim_{x \to 0} \frac{sin(sinx)-x(1-x^2)^{\frac{1}{3}}}{x^3}$

2. $\displaystyle \frac{\left(x-\frac{x^{3}}{3}+\frac{x^{5}}{10}-\frac{8x^{7}}{315}+\frac{13x^{9}}{2520}-............\right)-\left(x-\frac{x^{3}}{3}-\frac{x^{5}}{9}-\frac{5x^{7}}{81}-\frac{10x^{9}}{243}-......\right)}{x^{3}}$

$\displaystyle =\frac{19x^{2}}{90}+\frac{103x^{4}}{2835}+\frac{31 51x^{6}}{68040}+................$

$\displaystyle \lim_{x\to 0}\left[\frac{19x^{2}}{90}+\frac{103x^{4}}{2835}+\frac{315 1x^{6}}{68040}+................\right]=0$

3. Thanks , but how you know how todo Taylor's theorem on this term :
$\displaystyle -x(1-x^2)^\frac{1}{3}$
?

4. Originally Posted by gilyos
Thanks , but how you know how todo Taylor's theorem on this term :
$\displaystyle -x(1-x^2)^\frac{1}{3}$
?
You don't need to use Taylor's Theorem to do that

you should expand it by binomial theroem

$\displaystyle (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + $$\displaystyle \frac{n(n-1)(n-2)}{3!} x^3 + \frac{n(n-1)(n-2)(n-3)}{4!} x^4 +$$\displaystyle \frac{n(n-1)(n-2)(n-3)(n-4)}{5!}x^5 + ...$

In this case $\displaystyle n= 1/3 , x <-> -x^2$

$\displaystyle ( 1- x^2)^{1/3} = 1 - \frac{x^2}{3} + ....$

the next terms are left to you since I don't get a pencil and a piece of paper right now .

You will see they are exactly what galactus has written for us