$$\lim _{x\to 0}\left(\frac{1-\cos\left(2x\right)}{2x^2}\right)^{\frac{1}{x^2}}$$ I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks

I tried to solve like that:(Taylor) $$e^{\frac{1}{x^2}\ln\left(\frac{1-\cos\left(2x\right)}{2x^2}\right)}$$ $$\ln\left(\frac{1-\cos\left(2x\right)}{2x^2}\right)=\ln\left(\frac{1-1+2x^2-\frac{2x^4}{3}}{2x^2}\right)=\ln\left({1-\frac{2x^4}{9x^2}}\right)$$

$$\frac{1}{x^2}\left(-\frac{2x^4}{9x^2}\right)=\color{red}{e^{-\frac{2}{9}}}$$ Which however it is wrong. The result sought is $\color{red}{e^{-1/3}}$

  • 2
    $\begingroup$ $(2-2/3)/2=1-1/3$ $\endgroup$ – Peter Franek Jan 27 '16 at 22:06

You made a mistake at the second equality: how did $\frac{2x^2 - \frac{2x^4}{3}}{2x^2}$ become $1- \frac{2x^4}{9x^2}$? It should be a $6$, not a $9$ — and then you'll get the correct limit.

But besides this, be careful: for what you write to be correct, do not forget to include the $o(\cdot)$'s in the Taylor expansions! Without them, this is rigorously wrong, and also an open invitation to many mistakes (since it's not explicit to anyone, including you, what order becomes negligible.)

  • $\begingroup$ OMGGGGGGGGGGGGG :) Thanks $\endgroup$ – Amarildo Jan 27 '16 at 22:09
  • $\begingroup$ @ClementC., is there any sufficient method involving $$\frac{1-\cos(2x)}{2x^2}=\frac{\sin^2x}{x^2}?$$ $\endgroup$ – Invisible Jan 18 '20 at 11:57

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.