# Simple limit with asymptotic approach. Where's the error?

Simply calculus question about a limit.

I don't understand why I'm wrong, I have to calculate $$\lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}\sin x}{1 - \cos\sqrt{x^3}}$$

Using asymptotics, limits and De l'Hospital rule I would write these passages...

$$= \lim_{x \rightarrow 0} \frac{x \, (2 - \sqrt[3]{8 - x^2})}{x^3/2} = \lim_{x \rightarrow 0} \frac{\frac{2}{3}\frac{\sqrt[3]{8 - x^2}}{8-x^2}x}{x} = \frac{1}{6}$$

But the answers should be $\frac{5}{6}$. Thank you for your help.

• What is your question? Is it to find the error? Or to get $\frac56$ (this one is easy). – José Carlos Santos Aug 17 '18 at 14:31
• Maybe both but probably just to find and explain the error should be enough. Thanks – Marco Aug 17 '18 at 14:35
• In order to find the error, it would be necessary that you provide the details of your calculation. I don't know how you got the expression after "passages...", so how do you expect me to find the mistake? – A. Pongrácz Aug 17 '18 at 14:43
• In the first passage I've used $\sin x \sim x$ and $1 - \cos \sqrt{x^3} \sim \frac{x^3}{2}$. In the second passage I've used De l'Hospital rule after an $x$ simplification. – Marco Aug 17 '18 at 14:56
• The expression is not defined for $x<0,$ because of the $\sqrt {x^3}$ term. – zhw. Aug 17 '18 at 22:44

The mistake lies at the beginning : $$\lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}(x-\frac{x^3}{6})}{1 - \cos\sqrt{x^3}} =\frac56$$
$$\lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}\:x}{1 - \cos\sqrt{x^3}} =\frac16$$
At denominator $1-\cos(x^{3/2})$ is equivalent to $\frac12 x^3$. Thus one cannot neglect the $x^3$ terms in the numerator. So, the equivalent of $\sin(x)$ must not be $x$ but $x-\frac{x^3}{6}$ . This was the trap of the exercise.
You are just replacing $\sin x$ by $x$ and $\cos \sqrt{x^3}$ with $1-(x^3/2)$. Both these replacements are wrong for the very simple reason that $\sin x\neq x$ and $\cos \sqrt{x^3}\neq 1-(x^3/2)$ unless $x=0$.
The right approach is to use the standard limits $$\lim_{x\to 0}\frac{\sin x} {x} =1,\,\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}\tag{1}$$ Using above limit and L'Hospital's Rule it can be easily proved that $$\lim_{x\to 0}\frac{x-\sin x} {x^3}=\frac {1}{6}\tag{2}$$ Another limit which is needed here is $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{3}$$ We can evaluate the limit in question using limits $(1), (2)$ and $(3)$ as follows \begin{align} L&=\lim_{x\to 0}\frac{2x-\sin x\sqrt[3]{8-x^2}}{1-\cos\sqrt{x^3}}\notag\\ &= \lim_{x\to 0}\frac{2x-\sin x\sqrt[3]{8-x^2}}{x^3}\cdot\frac{x^3}{1-\cos\sqrt{x^3}}\notag\\ &= 2\lim_{x\to 0}\frac{2x-2\sin x+\sin x(2-\sqrt[3]{8-x^2}) } {x^3}\text{ (using (1))}\notag\\ &=2\left(\frac{1}{3}+\lim_{x\to 0}\frac{\sin x} {x} \cdot\frac{2-\sqrt[3]{8-x^2}}{x^2}\right)\text{ (using (2))}\notag\\ &= \frac{2}{3}+2\lim_{t\to 8}\frac{8^{1/3}-t^{1/3}}{8-t} \text{ (putting }t=8-x^2) \notag\\ &= \frac{2}{3}+2\cdot\frac{1}{3}\cdot 8^{-2/3}\text{ (using (3))}\notag\\ &=\frac{5}{6}\notag \end{align}