i tried to solve it

[math]

\lim _{x->0} \frac{cos(xe^x)-cos(xe^{-x})}{x^3}\\

[/math]

[math]

e^x=1+x+O(x^2)\\

[/math]

[math]

e^{-x}=1-x+O(x^2)\\

[/math]

[math]

xe^x=x+x^2+O(x^2)

[/math]

[math]

xe^{-x}=x-x^2+O(x^2)

[/math]

[math]

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

[/math]

[math]

\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}=\\

[/math]

[math]

=\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\\

[/math]

 

the answer is 1/2

why i got 0??

You didn't carry enough terms. Suggestion: Try expanding the exponential after doing the cosine expansion. In other words, use [math]\cos(xe^x)\approx 1-\frac 1 2 (xe^x)^2 + \cdots[/math]

 

BTW, the answer is neither zero nor 1/2.