April 8th, 2009, 09:40 AM  #1 
Newbie Joined: Apr 2009 Posts: 2 Thanks: 0  taylor series for limit
I am supposed to find the limit as x > 0 of (e^(x^2)  1  x^2) / (x sinx  x^2) I think that I am supposed to use Taylor Series, but I am not quite sure how to do that. Do I try to do a taylor series of the entire function? Or is there some way to break it down? 
April 8th, 2009, 01:33 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 723  Re: taylor series for limit
Expand the numerator and denominator separately and keep the lowest nonzero power of x for both. For the numerator you need to get Taylor series for e^(x^2) while for the denominator you need to get series for sinx. For both the surviving term will be some constant times x^4, so your answer is the ratio of the constants.

April 8th, 2009, 02:54 PM  #3 
Newbie Joined: Apr 2009 Posts: 2 Thanks: 0  Re: taylor series for limit
ok, I'm not sure if I followed your instructions correctly, but what I came up with is: for the numerator, I found the taylor series for e^(x^2): 1 + x^2 + x^4/2! + x^6/3! ... then I added in the rest of the numerator (1  x^2) to get: x^4/2 ... for the denominator, I found the taylor series for sinx and then added / multiplied in the rest of the denominator to get: x^4/6 + x^6/5! x^8/7! .... taking the ratio of the x^4 terms give me 1/2 * 6/1 = 3 is this correct? if so, is it just normal practice to ignore the rest of the terms in the series and just focus on the first one? Thanks!! 
April 9th, 2009, 03:57 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 723  Re: taylor series for limit
Your answer looks right. For your general question, dividing all the terms in the numerator and deominator by the same power of x leaves the value unchanged. When you take out the x^4 term you have for num. and den. a constant plus powers of x. The powers of x go to 0 as x goes to 0, so the method works. There is only one thing you have to be careful of in that both power series have to converge in some interval around x=0. This happens to be true here because the exp and sin functions are both well behaved. 

Tags 
limit, series, taylor 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
In need of help disk, series test, taylor, and power series  g0bearmon  Real Analysis  2  May 22nd, 2012 12:10 PM 
Taylor Series of log(1t)  vjj  Real Analysis  2  April 5th, 2012 10:40 AM 
Need help on Taylor's series limit calculation  dttah  Calculus  3  November 23rd, 2011 02:24 PM 
Help with Taylor Series  helpatmath  Real Analysis  1  April 26th, 2009 03:46 PM 
In need of help disk, series test, taylor, and power series  g0bearmon  Calculus  1  December 31st, 1969 04:00 PM 