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In my notes it was given that as $x\rightarrow 0,$

$$\frac{x^{3/2}}{1-\cos x} = \mathcal{O}(x^{-1/2})$$

It didnt give any explanation, so I was wondering what would the "method"/"intuition" be behind it, to be able to deduce/find out that $\mathcal{O}(x^{-1/2})?$

And how would I then find the order of $$\frac{x^{3/2}}{1+\sin x}$$ as $x\rightarrow 0.$

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I have made a change to the tag, then cancelled the change. Question's text is intact. –  Emmad Kareem Feb 22 '12 at 23:35

1 Answer 1

up vote 2 down vote accepted

The Taylor series for $1-\cos x$ begins with a term in $x^2$, and $x^{3/2}/x^2=x^{-1/2}$. You can do the same approach for your $\sin x$ example.

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Thanks, just to make sure, the taylor series for $1+sinx$ starts like $1+x-\frac{x^3}{6}$ etc. I SHOULDNT ignore the 1, correct? i.e. I can take $x^{3/2}/1=x^{3/2}$? –  Heijden Feb 22 '12 at 23:34
    
That's correct. –  Gerry Myerson Feb 22 '12 at 23:39
    
Okay, thanks for your help Gerry. :) –  Heijden Feb 22 '12 at 23:49

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