Hello everyone, please help me with this problem.
$\displaystyle \lim _{x\to 0}\frac{x}{\sin ^3 x}-\frac{\sin x}{x^3}.$
I will appreciate any help. Thank you very much.
$\displaystyle \frac{x}{(\sin(x))^3}-\frac{\sin(x)}{x^3}=\frac{x^4-(\sin(x))^4}{(\sin(x))^3x^3}$
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
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.
I have the same impression.(P.P.S. I'm sure that the above quoted text is not what it said when I started answering this question!)
Thank you very much. That idea never ocurred to me; I will work in that way too.
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about the p.p.s ;
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.