0
$\begingroup$

$$\lim _{x\to 0}\left(\frac{\cos \left(xe^x\right)-\cos \left(xe^{-x}\right)}{x^3}\right)=?$$

Trig transformations and such don't seem to help at all, and L'Hospitals rule just complicates matters even more. So how would you solve this limit, thanks.

$\endgroup$
  • $\begingroup$ By the way, it may be long but using L'Hospital three times does give the answer. $\endgroup$ – Vincenzo Oliva Jan 27 '16 at 8:29
1
$\begingroup$

Since $\cos t=1-t^2/2+o(t^3)$, we have $$ \cos(xe^x)-\cos(xe^{-x})= -\frac{1}{2}x^2e^{2x}+\frac{1}{2}x^2e^{-2x}+o(x^3) $$ Now it should be quite easy.

$\endgroup$
5
$\begingroup$

Without Taylor:

By the difference of two cosines,

$$\frac{\cos(xe^x)-\cos(xe^{-x})}{x^3}=-2\frac{\sin(x\cosh(x))\sin(x\sinh(x))}{x^3}.$$

Then

$$-2\frac{\sin(x\cosh(x))\sin(x\sinh(x))}{x^3}=-2\frac{\sin(x\cosh(x))}{x\cosh(x)}\frac{\sin(x\sinh(x))}{x\sinh(x)}\cosh(x)\frac{\sinh(x)}x$$

which tends to $-2\cdot1\cdot1\cdot1\cdot1.$

$\endgroup$
  • $\begingroup$ Elegant, for sure ! $\endgroup$ – Claude Leibovici Jan 27 '16 at 8:55
  • $\begingroup$ @ClaudeLeibovici: I was too lazy to compute the Taylor of the composed function :) $\endgroup$ – Yves Daoust Jan 27 '16 at 9:01
1
$\begingroup$

$\cos(xe^x) = 1 - \dfrac{x^2e^{2x}}{2} + \dfrac{x^4e^{4x}}{24} + O(x^6)$, and $\cos(xe^{-x}) = 1-\dfrac{x^2e^{-x}}{2}+\dfrac{x^4e^{-4x}}{24} + O(x^6)$. You can take it from here.

$\endgroup$
1
$\begingroup$

Taylor series would help $$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^5\right)$$ $$\cos(xe^x)=1-\frac 12 xe^x+\frac 1{24} (xe^x)^2+\cdots$$ $$xe^x=x+x^2+\frac{x^3}{2}+\frac{x^4}{6}+O\left(x^5\right)$$ $$\cos(xe^x)=1-\frac{x^2}{2}-x^3-\frac{23 x^4}{24}+O\left(x^5\right)$$ Do the same for the second term.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.