The problem is:
f(x) = [cos(3x^2)-1]/(x^2) at x=0
I know that the Mclaurin series for cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ....
That is where I'm stuck, I can't remember how to use the Mclaurin series with this function
The problem is:
f(x) = [cos(3x^2)-1]/(x^2) at x=0
I know that the Mclaurin series for cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ....
That is where I'm stuck, I can't remember how to use the Mclaurin series with this function