Solving limit using tailor series

[transgalactic]

<br /> \lim _{x-&gt;0} \frac{cos(xe^x)-cos(xe^{-x})}{x^3}\\<br />
<br /> e^x=1+x+O(x^2)\\<br />
<br /> e^{-x}=1-x+O(x^2)\\<br />
<br /> xe^x=x+O(x^2)\\<br />
<br /> cos(x)=1-\frac{1}{2!}x^2+O(x^3)\\<br />
<br /> \lim_{x-&gt;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}<br />

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
<br /> \lim _{x-&gt;0} \frac{cos(xe^x)-cos(xe^{-x})}{x^3}\\<br /> e^x=1+x+O(x^2)\\<br /> &lt;br /&gt; e^{-x}=1-x+O(x^2)\\&lt;br /&gt;<br /> &lt;br /&gt; xe^x=x+x^2+O(x^2)&lt;br /&gt;<br /> &lt;br /&gt; xe^{-x}=x-x^2+O(x^2)&lt;br /&gt;<br /> &lt;br /&gt; cos(x)=1-\frac{1}{2!}x^2+O(x^3)\\&lt;br /&gt;<br /> &lt;br /&gt; \lim_{x-&amp;gt;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}=\\&lt;br /&gt;<br /> &lt;br /&gt; =\lim_{x-&amp;gt;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\\&lt;br /&gt;<br /> <br /> the answer is 1/2<br /> 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? )