roman

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??

there is a function which is differentiable continuously on [a,b] . suppose |f'(x)|<1 for all[math] x\epsilon [a,b][/math] prove that there exists 0<=k<1 [math] x_1\epsilon [a,b] [/math] [math] x_2\epsilon [a,b] [/math] that this equation is true [math] |f(x_1)-f(x_2)|<=K|x_1-x_2| [/math]