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I'm trying to get my head around this equation,

$$\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{1-\cos x}$$

but nothing I do seems to make it any more clearer. Do any one know how to do it?

egreg
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Salviati
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  • Did you try L'Hospital ? The limit is $-\frac{4}{3}$. If you use L'Hospital, you will have to differentiate numerator and denominator twice. – Peter Apr 19 '16 at 15:42
  • Yes, but I am not that familiar with it yet so I must be doing it wrong. We will go through it in class next week, I'm just trying to be prepared. – Salviati Apr 19 '16 at 15:44
  • Yes I refreshed now and saw it. Will try it, thank you. – Salviati Apr 19 '16 at 15:51

3 Answers3

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\begin{align} \lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{1-\cos x} &=\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{x^2}\cdot\frac{x^2}{1-\cos x} \\ &=\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{x^2}\cdot\lim_{x\to 0}\frac{x^2}{1-\cos x}\\ &=\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{x^2}\cdot2 \\ &=2\lim_{x\to 0}\frac{2x-\frac{8x^3}{3!}+O(x^5)-2x-\frac23x^2-O(x^3)}{x^2}\\ &=2\lim_{x\to 0}\frac{-\frac{2x^2}{3}+O(x^5)-O(x^3)}{x^2} \\ &=-\frac{4}3 \end{align}

GoodDeeds
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  • @Uti But I would suggest you try the solution via L'Hospital as an exercise. – Peter Apr 19 '16 at 16:21
  • @Peter Yes I will. Still trying to figure out how it work dough, the f'(x) / g'(x) = 0 part gets me a bit confused. – Salviati Apr 19 '16 at 16:23
  • This means that you can apply L'Hospital to this quotient. If you differentiate again, you are done because now you can simply insert $x=0$. – Peter Apr 19 '16 at 16:32
  • Ahh I finally got it! Sorry, language was a barrier, used Google Translate. I will post "my own" solution also :) Thank you! – Salviati Apr 19 '16 at 16:53
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Use Taylor expansion: $$ 1-\cos x=1-1+\frac{x^2}{2}+o(x^2)=\frac{x^2}{2}+o(x^2) $$ so you need to go to degree $2$ also in the numerator.

Use $\sin2x=2x+o(x^2)$ and $$ \sqrt[3]{1+x}=1+\frac{x}{3}+o(x) $$ so you have $$ \lim_{x\to0}\frac{2x+o(x^2)-2x(1+x/3+o(x))}{x^2/2+o(x^2)}= \lim_{x\to0}\frac{-2x^2/3+o(x^2)}{x^2/2+o(x^2)}=-\frac{4}{3} $$

egreg
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Using L'Hôpital's Rule $$\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{1-\cos x}$$

Try x = 0

$$\lim_{x\to 0}\frac{\sin0-2*0(1)^{1/3}}{1-\cos 0} = \frac{0}{0}$$ Derive f(x) and g(x) and set x = 0

$$\frac{f'(x)}{g'(x)}=\lim_{x\to 0}\frac{2\cos2x-\frac{2(4x+3)}{3(x+1)^{2/3}}}{-\sin x}= \frac{0}{0}$$ Derive again and set x = 0 $$\frac{f''(x)}{g''(x)}=\lim_{x\to 0}\frac{-4\sin2x-\frac{4(2x+3)}{9(x+1)^{5/3}}}{\cos x}= -\frac{4}{3}$$

Salviati
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