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April 8th, 2009, 09:40 AM   #1
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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?
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April 8th, 2009, 01:33 PM   #2
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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.
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April 8th, 2009, 02:54 PM   #3
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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!!
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April 9th, 2009, 03:57 PM   #4
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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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