# Series expansions

1. Sep 22, 2009

### Physicist_FTW

use series expansion to fine the limit as x--->0 of

1/(sin^2)(x) - 1/x^2

Sin^2(x)=(X-X^3/3!)^2 ive assumed this gave me
Sin^2(x)=X^2-2X^4/6+X^6/36
flip this over is that equivalent to 1/Sin^2(x)?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 22, 2009

### Dick

It's not totally equivalent because you omitted higher order terms. Can you show you have enough terms to capture the behavior of 1/sin^2(x)-1/x^2 in the sense that terms that you omitted will go to zero as x->0?

3. Sep 23, 2009

### Physicist_FTW

Im sorry im not quite sure what you mean by that?

4. Sep 23, 2009

### Dick

I'm just saying sin(x) isn't equal to x-x^3/3!. You omitted the higher order terms, like x^5/5!. As you work out the limit you'll want to convince yourself that including them doesn't affect the limit as x->0. Do you know how to find the limit of 1/(x-x^3/3!)^2-1/x^2?

5. Sep 23, 2009

### Physicist_FTW

Actually no i dont, maybe you could explain to me how?

6. Sep 23, 2009

### Dick

Factor the denominator of x-x^3/6 as x*(1-x^2/6). So you've got (1/x)*(1/(1-x^2/6)). Use that 1/(1-a)=1+a+a^2+a^3+... (the usual geometric series thing) to move the second factor into the numerator.
Now you've got (1/x)^2*(1+x^2/6+...)^2-1/x^2. Expand it. Now go back and figure out why I didn't need to keep any higher powers of x than I did.