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 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: 6,759 Thanks: 696 Re: taylor series for limit Expand the numerator and denominator separately and keep the lowest non-zero 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: 6,759 Thanks: 696 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.

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