Maclaurin series...?

Question

Find the limit of 1-cos(3x/2)/(1+x-e^x) as x goes to 0, by using Maclaurin series for 1-cos(3x/2) and 1+x-e^x

oh that should be

[1-cos(3x/2)] / (1+x-e^x)

Answer 1

The Maclaurin series for cos(t) is

cos(t) = 1 - t²/2! + (t^4)/4! - + ...

so

cos(3x/2) = 1 - (3x/2)²/2! + [(3x/2)^4]/4! -+...

and 1 - cos(3x/2) = (3x/2)²/2! - [(3x/2)^4]/4! + -...

e^x = 1 + x + (x²)/2! + (x³)/3! + ...

so

1 + x - e^x = - ((x²)/2! + (x³)/3! + ...)