Solving limit using tailor series

[transgalactic]

$$\lim _{x->0} \frac{cos(xe^x)-cos(xe^{-x})}{x^3}\\$$
$$e^x=1+x+O(x^2)\\$$
$$e^{-x}=1-x+O(x^2)\\$$
$$xe^x=x+O(x^2)\\$$
$$cos(x)=1-\frac{1}{2!}x^2+O(x^3)\\$$
$$\lim_{x->0} \frac{1-\frac{1}{2!}(x+O(x^2))^2+O(x^3)-1+\frac{1}{2!}(x+O(x^2))^2+O(x^3)}{x^3}$$

but i don't know how to deal with the remainders
there squaring of them etc..

??

 

[tiny-tim]

Hi transgalactic!

hmm … from the x³ on the bottom, I'd guess you need to specify the x² terms as well, and not start the Os until O(x³)

or you could just use cosA - cosB = 2.sin(A+B)/2.sin(A-B)/2

 

[transgalactic]

i substituted functions one into another
that what i got.
where is my mathematical mistake there??

how to open this expression and having one O()
??

 

[transgalactic]

i tried to solve it again
[tex]
\lim _{x->0} \frac{cos(xe^x)-cos(xe^{-x})}{x^3}\\
e^x=1+x+O(x^2)\\
$$e^{-x}=1-x+O(x^2)\\$$
$$xe^x=x+x^2+O(x^2)$$
$$xe^{-x}=x-x^2+O(x^2)$$
$$cos(x)=1-\frac{1}{2!}x^2+O(x^3)\\$$
$$\lim_{x->0} \frac{1-\frac{1}{2!}(x+x^2+O(x^2))^2+O(x^3)-1+\frac{1}{2!}(x-x^2+O(x^2))^2+O(x^3)}{x^3}=\\$$
$$=\lim_{x->0} \frac{1-\frac{1}{2!}(x^2+O(x^2))+O(x^3)-1+\frac{1}{2!}(x^2+O(x^2))+O(x^3)}{x^3}=0\\$$

the answer is 1/2
why i got 0??

 

[tiny-tim]

Hi transgalactic!

Thta's actually pretty good, except …

$$cos(x)=1-\frac{1}{2!}x^2+O(x^3)\\$$

should be O(x⁴) at the end, not O(x³)

and in the last line you should have x³ terms also …

they're the ones that don't cancel!

(and btw, it's "taylor", and why do you keep writing 2! instead of just 2? )