# Thread: Find limit using taylor series

1. ## 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!!

2. Hi

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
$\frac{\frac{x^4}{2}+\frac{x^6}{6}}{-\frac{x^4}{6}+\frac{x^6}{120}} = \frac{\frac{x^4}{2}+\frac{x^6}{6}}{-\frac{x^4}{6}\left(1-\frac{x^2}{20}\right)} = \frac{-3-x^2}{1-\frac{x^2}{20}}$
which gives $(-3-x^2)\left(1+\frac{x^2}{20}\right) = -3-\frac{23}{20}\:x^2$