Hi
Your work is OK
You have found an equivalent to 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
which gives
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!!