# using a series to evaluate a limit:

• May 7th 2009, 08:54 AM
ryu991
using a series to evaluate a limit:
How would I change this problem into a series and evaluate the limit?

(e^x) - 1
--------
sinx

any help greatly appreciated as I'm completely stuck.
• May 7th 2009, 09:11 AM
HallsofIvy
The MacLaurin series for [tex]e^x is 1+ x+x^2/2+ x^3/3!+ ...
What do you get when you subtract 1 from that.

The MacLaurin series for sin(x) is x- x^3/3!+ x^5/5!+ .... If you divide the previous series by that, do you see that you can cancel an "x"? Then, as x goes to 0, all but the constant terms will disappear.
• May 7th 2009, 09:49 AM
ryu991
Quote:

Originally Posted by HallsofIvy
The MacLaurin series for [tex]e^x is 1+ x+x^2/2+ x^3/3!+ ...
What do you get when you subtract 1 from that.

The MacLaurin series for sin(x) is x- x^3/3!+ x^5/5!+ .... If you divide the previous series by that, do you see that you can cancel an "x"? Then, as x goes to 0, all but the constant terms will disappear.

ok, so the first x's cancel and then as I divide. (x^2/2! divided by -x^3/3! etc...)

I wind up with:

-3x^-1 + 20x ^-2 - 210x^ -3 etc....

So 2 things:

1) is my division right? there is clearly a pattern just not sure if its the right one.

2) If it is right do I just need to take the derivative of each and calculate the limit from there or is there more i'm missing?

edit: as x goes to 0 how would all the x's dissapear. Wouldnt I just get a bunch of 0 over 0's?
• May 7th 2009, 09:58 AM
Spec
Big O notation - Wikipedia, the free encyclopedia

$\displaystyle \lim_{x \to 0}\frac{x+O(x^2)}{x+O(x^3)}=\lim_{x \to 0}\frac{1+O(x)}{1+O(x^2)}=1$
• May 7th 2009, 10:06 AM
ryu991
Quote:

Originally Posted by Spec
Big O notation - Wikipedia, the free encyclopedia

$\displaystyle \lim_{x \to 0}\frac{x+O(x^2)}{x+O(x^3)}=\lim_{x \to 0}\frac{1+O(x)}{1+O(x^2)}=1$

I'm sorry I dont understand what that means. I tried the wiki link but that just confused me more.
• May 7th 2009, 10:21 AM
Spec
Instead of messing around with all the $\displaystyle c_2x^2+c3x^3 + ... + c_nx^n$ terms, you can simply say that $\displaystyle e^x = 1+ x+O( x^2)$ with an error smaller than $\displaystyle |kx^2|$ as $\displaystyle x \to 0$, where $\displaystyle k$ is a constant. In particular, $\displaystyle \lim_{x \to 0}O(x^n)=0$ for all $\displaystyle n > 0$.
• May 7th 2009, 10:25 AM
ryu991
Quote:

Originally Posted by Spec
Instead of messing around with all the $\displaystyle c_2x^2+c3x^3 + ... + c_nx^n$ terms, you can simply say that $\displaystyle e^x = 1+ x+O( x^2)$ with an error smaller than $\displaystyle |kx^2|$ as $\displaystyle x \to 0$, where $\displaystyle k$ is a constant. In particular, $\displaystyle \lim_{x \to 0}O(x^n)=0$ for all $\displaystyle n > 0$.

the problem with that is my professor specifically wants us to use a series to evaluate the limit so I wouldnt be allowed to do it that way. thanks for the suggesstion though.
• May 7th 2009, 10:36 AM
Spec
It's basically the same thing. I have just truncated the Maclaurin series for $\displaystyle e^x$ and $\displaystyle \sin x$ a bit so you don't have to write out all the terms. You can of course still write it like this:

$\displaystyle \lim_{x \to 0}\frac{(1 + x + \frac{x^2}{2!} + ... + \frac{x^{n}}{n!})-1}{x+\frac{x^3}{3!}+...+\frac{x^{2n+1}}{(2n+1)!}}= \lim_{x \to 0}\frac{1 + \frac{x}{2!} + ... + \frac{x^{n-1}}{n!}}{1+\frac{x^2}{3!}+...+\frac{x^{2n}}{(2n+1) !}}=1$
• May 7th 2009, 10:38 AM
ryu991
I think I got the first part, I write out all the terms in the form you did, but I'm confused as to what to do from there.

Why does it all equal to 1?
• May 7th 2009, 10:58 AM
Spec
All the x terms vanish as x goes to zero.
• May 7th 2009, 11:01 AM
ryu991
Quote:

Originally Posted by Spec
All the x terms vanish as x goes to zero.

1st off thanks for all the help.

2nd, and this is the same question I asked the other guy..

why do all the x terms vanish. As x approaches 0 wouldnt we be stuck with a bunch of 0 over 0's?

I know the limit is 1 because I already did it with l' hospital but I just dont understand how the x's vanish and why the series does not wind up in an inderteminate form.
• May 7th 2009, 11:34 AM
Spec
No, because you have $\displaystyle \frac{1 + ...}{1+...}$ where $\displaystyle ...$ is a sum of x terms that goes to zero as x goes to zero. So the limit becomes $\displaystyle \frac{1+0}{1+0}=1$
• May 7th 2009, 11:44 AM
ryu991
Quote:

Originally Posted by Spec
No, because you have $\displaystyle \frac{1 + ...}{1+...}$ where $\displaystyle ...$ is a sum of x terms that goes to zero as x goes to zero. So the limit becomes $\displaystyle \frac{1+0}{1+0}=1$

ahhhh, that was so obvious I cant believe I didnt notice that. Thanks so much for all your help.