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Prove $$ \lim_{x \to 0}\frac{x^3 - \sin^3x}{x - \ln{(1+x)} - 1 + \cos x} = 0 $$

Obviously, Using l'Hôpitals rule, we can evaluate this limit. But, taking derivatives of such functions is such a mess. Anyone sees a trick to do this problem faster? any ideas?

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Such a mess? Is not a mess. The most you have to use the chain rule once! –  Pragabhava Oct 16 '12 at 2:18
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But if you are really too lazy to do it, a quick example would be $\sin x \sim x$ for small $x$. What does that say of $\sin^3 x$? And so on. –  Pragabhava Oct 16 '12 at 2:24

2 Answers 2

up vote 3 down vote accepted

$$\dfrac{x^3 - \sin^3(x)}{x - \ln(1+x) - 1 + \cos(x)} = \dfrac{x - \sin(x)}{x} \times \dfrac{x^2 + x \sin(x) + \sin^2(x)}{1 - \dfrac{\ln(1+x)}{x}- \dfrac{1-\cos(x)}{x}}$$

$$1 - \dfrac{\ln(1+x)}{x}- \dfrac{1-\cos(x)}{x} = 1 - \left(1 - \dfrac{x}2 +\dfrac{x^2}3 - \cdots \right) - \left( \dfrac{\dfrac{x^2}{2!} - \dfrac{x^4}{4!} + \cdots}{x} \right) \\ = \left(\dfrac{x}2 - \dfrac{x^2}3 + \cdots \right) - \left(\dfrac{x}2 - \dfrac{x^3}{4!} + \cdots \right) = -\dfrac{x^2}3 + \mathcal{O}(x^4)$$

$$\dfrac{x - \sin(x)}{x} = \dfrac{x^2}{3!} + \mathcal{O}(x^4)$$

Hence,$$\dfrac{x^3 - \sin^3(x)}{x - \ln(1+x) - 1 + \cos(x)} = \left(\dfrac{x^2}{3!} + \mathcal{O}(x^4) \right) \times \dfrac{x^2 + x \sin(x) + \sin^2(x)}{-\dfrac{x^2}3 + \mathcal{O}(x^4)}$$

Now you should be able to finish it off.

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Dear Sir. Thank you for your nice solution. Our proof below is more elementary since we do not use Taylor's expansion. –  blindman Oct 16 '12 at 2:50

We have $x^3-\sin^3x=(x-\sin x)(x^2+x\sin x+\sin^2x)\approx\frac{x^3}{6}(3x^2)=\displaystyle\frac{x^5}{2}$ as $x\to 0$. Hence \begin{equation*} \begin{array}{lll} \displaystyle\lim_{x \to 0}\frac{x^3 - \sin^3x}{x - \ln{(1+x)} - 1 + \cos x} &=&\displaystyle\lim_{x\to 0}\frac{x^5}{2(x - \ln{(1+x)} - 1 + \cos x)}\\ &\overset{H}{=}&\displaystyle\lim_{x \to 0}\frac{5x^4}{2(1-\frac{1}{1+x}-\sin x)}\\ &\overset{H}{=}&\displaystyle\lim_{x \to 0}\frac{10x^3}{\frac{1}{(1+x)^2}-\cos x}\\ &\overset{H}{=}&\displaystyle\lim_{x \to 0}\frac{30x^2}{\frac{-2}{(1+x)^3}+\sin x}\\ &=&\displaystyle\frac{0}{\frac{-2}{(1+0)^3}+0}=0. \end{array} \end{equation*}

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