Order of trig functions as $x \rightarrow 0$

Question

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 $x\rightarrow 0.$

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

score 2 · Accepted Answer · answered Feb 22 '12 at 23:18

2

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.

answered Feb 22 '12 at 23:18

Gerry Myerson

179,216

1

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

Order of trig functions as $x \rightarrow 0$

1 Answers1