1. ## limit

$\displaystyle \lim _{x\to 0}\frac{x}{\sin ^3 x}-\frac{\sin x}{x^3}.$

I will appreciate any help. Thank you very much.

2. ## Re: limit

Using $\displaystyle \sin x\sim x\;(x\to 0)$ we can express $\displaystyle \dfrac{x}{\sin^3 x}-\dfrac{\sin^3 x}{x^3 }\sim \dfrac{x^4-\sin^6 x}{ x^6}$

3. ## Re: limit

Thank you for your quick response, but my problem is

$\displaystyle \dfrac{x}{\sin^3 x}-\dfrac{\sin x}{x^3}$

, not $\displaystyle -\dfrac{\sin ^3 x}{x^3}.$

4. ## Re: limit

Originally Posted by joll
Thank you for your quick response, but my problem is

$\displaystyle \dfrac{x}{\sin^3 x}-\dfrac{\sin x}{x^3}$

, not $\displaystyle -\dfrac{\sin ^3 x}{x^3}.$

$\displaystyle \frac{x}{(\sin(x))^3}-\frac{\sin(x)}{x^3}=\frac{x^4-(\sin(x))^4}{(\sin(x))^3x^3}$

Which is indeterminate so we can do a number of things, we could use L'Hopital's rule (twice?) or expand top and bottom as power series about x=0 or ...

However we do this we get a finite limit.

(P.S. you are allowed to use a calculator or some other mechanical assistance to get some idea of what the limit might be, which provides a useful cross check on your analysis)

(P.P.S. I'm sure that the above quoted text is not what it said when I started answering this question!)

CB

5. ## Re: limit

Originally Posted by FernandoRevilla
Using $\displaystyle \sin x\sim x\;(x\to 0)$ we can express $\displaystyle \dfrac{x}{\sin^3 x}-\dfrac{\sin^3 x}{x^3 }\sim \dfrac{x^4-\sin^6 x}{ x^6}$

$\displaystyle \dfrac{x}{\sin^3 x}-\dfrac{\sin x}{x^3 }\sim \dfrac{x^4-\sin^4 x}{ x^6}$

CB

6. ## Re: limit

Originally Posted by CaptainBlack
$\displaystyle \dfrac{x}{\sin^3 x}-\dfrac{\sin x}{x^3 }\sim \dfrac{x^4-\sin^4 x}{ x^6}$
Perhaps, I read another thing. Now we can express $\displaystyle \dfrac{x^4-\sin^4 x}{ x^6}=\dfrac{x^2+\sin^2 x}{x^3}\cdot \dfrac{x^2-\sin^2 x}{x^3}$ and the limit is easier to find.

(P.P.S. I'm sure that the above quoted text is not what it said when I started answering this question!)
I have the same impression.

7. ## Re: limit

Thank you for your quick response.
Actually, I've tried L'Hospital's rule, but not twice. I will try that way first, thank you very much.

8. ## Re: limit

Thank you very much. That idea never ocurred to me; I will work in that way too.

----

I am terribly sorry for the confusion; actually, I didn't realize the mistake, and after I posted my reply saying

$\displaystyle \dfrac{x^4-(\sin x)^6}{x^6} = \dfrac{1}{x^2} - \left( \dfrac{\sin x}{x} \right)^6 \to \infty$,

I realized it was mistaken and totally editted my post. Perhaps I should have posted another reply. I am sorry.

9. ## Re: limit

Originally Posted by joll
I realized it was mistaken and totally editted my post. Perhaps I should have posted another reply. I am sorry.
It does not matter.