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$$\lim _{x\to 0}\left(\frac{1-\cos\left(2x\right)}{2x^2}\right)^{\frac{1}{x^2}}$$ 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 $\color{red}{e^{-1/3}}$

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

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  • $\begingroup$ OMGGGGGGGGGGGGG :) Thanks $\endgroup$ – Amarildo Jan 27 '16 at 22:09

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