roman Posted March 10, 2009 Posted March 10, 2009 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??
D H Posted March 10, 2009 Posted March 10, 2009 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.
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