How can I verify that $\lim_{x \to 0} \dfrac{\cos{2x} - \sqrt{1-4x^2}}{2x \sin x^3} = \frac{4}{3}$?

I've tried with Taylor so that:

$2x\sin x^3 \sim 4x^4 - \frac{2}{3}x^{10}$

$\cos2x \sim 1- 2x^2$

$\sqrt{1-4x^2} \sim 1-2x^2-2x^4$

But it keeps me giving the wrong result, so the question is, am I using wrong the Taylor series?

When approximating the function and plugging them in the limit should them all be of the same order?

A well-detailed explanation is more than welcome.

  • $\begingroup$ Since the Taylor expansion of $\sin x=x-\frac{x^3}{6}+\cdots$, shouldn't $2x\sin x \sim 2x^4-\frac{x^{10}}{3}$? $\endgroup$ – Tim Thayer Oct 14 '17 at 16:24

Hint. First of all $2x\sin(x)^3=2x^4+o(x^4)$. Moroever, here you need a longer expansion for $\cos(2x)$: $$\cos(2x)=1-\frac{(2x)^2}{2}+\frac{(2x)^4}{4!}+o(x^4)=1-2x^2+\frac{2x^4}{3}+o(x^4).$$ Hence $$\frac{\cos(2x) - \sqrt{1-4x^2}}{2x \sin x^3}= \frac{1-2x^2+\frac{2x^4}{3} - (1-2x^2-2x^4)+o(x^4)}{2x^4+o(x^4)} $$ Do you mind to give another try?

P.S. I suggest the use of the little-o notation. It will help you to understand whether a Taylor expansion is sufficient for the purpose.

  • $\begingroup$ Thank you so much. But how can you tell if I need a longer expansion? $\endgroup$ – Ricouello Oct 14 '17 at 16:46
  • $\begingroup$ With the shorter expansion we have that $$\frac{\cos(2x) - \sqrt{1-4x^2}}{2x \sin x^3}= \frac{1-2x^2+o(x^2) - (1-2x^2-2x^4)+o(x^4)}{2x^4+o(x^4)}=\frac{o(x^2)}{2x^4+o(x^4)}$$ which is indeterminate. See the the little-o notation properties: en.wikipedia.org/wiki/Big_O_notation#Little-o_notation $\endgroup$ – Robert Z Oct 14 '17 at 16:52

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