Let's see:
To see the error using 2nd order approximation (since you have P2), we compute f''' (x)/3! * (x-a)^3.
In your case, a=0 as it's centered at a=0, and x=1/2 since we are estimating the error of f(1/2) from P2(1/2).
So we have f'''(1/2)/6 *(1/2)^3 = f'''(1/2)/48.
Since f'''(1/2) < 1, f'''(1/2)/48 < 1/48 as required.