# Find limit using taylor series

• Apr 9th 2009, 06:35 AM
generic1
Find limit using taylor series
I am supposed to find the limit as x -> 0 of (e^(x^2) - 1 - x^2) / (x sinx - x^2).

I do not know if my process or answer is at all correct, but here is what I have done:

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 and 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!!
• Apr 9th 2009, 06:49 AM
running-gag
Hi

Your work is OK (Clapping)

You have found an equivalent to $\frac{e^{x^2} - 1 - x^2}{x \sin x - x^2}$ in the neighborhood of 0 at the first order which is the same as the limit of the function at 0

If you want to find an equivalent at the second order

$\frac{\frac{x^4}{2}+\frac{x^6}{6}}{-\frac{x^4}{6}+\frac{x^6}{120}} = \frac{\frac{x^4}{2}+\frac{x^6}{6}}{-\frac{x^4}{6}\left(1-\frac{x^2}{20}\right)} = \frac{-3-x^2}{1-\frac{x^2}{20}}$

which gives $(-3-x^2)\left(1+\frac{x^2}{20}\right) = -3-\frac{23}{20}\:x^2$