,The right way to determine the slew-rate limted bandwidth, or the power bandwidth, as it's called, is to find the limiting effect of the slew-rate.
Slew-rate, SR, is the maximum possible rate of change of voltage per unit time:
SR=dv/dt.
If you have a sinusoidal signal of amplitude Vpk, you can write it like this:
v=Vpk*sinωt.
The voltage has the highest rate of change around 0, Π, 2Π and so on. This rate of change is found by differentiation:
dv/dt=Vpk*ω*cosωt. Its maximum occurs then for t=0, Π,...
For t=0 we get: dv/dt=Vpk*ω=Vpk*2*Π*f
This is the maximum rate of change of voltage, for a signal of amplitude Vpk and frequency f.
So, this rate of change must be lower than the slew-rate of the opamp, else the opamp will not be able to follow and distortion will occur.
In the limit, SR=Vpk*2*Π*fmax, where fmax is the maximum frequency of a signal of amplitude Vpk that an opamp having slew-rate SR will amplify without distortion:
fmax=SR/(Vpk*2*Π).
In your case, SR=2V/us and Vpk is 0.5V, so fmax=0.63MHz (for 1Vpp). That means at 1Vpp you can amplify a signal of only 0.63MHz. That is amolst 10 times less than your unity gain bandwidth, although the gain is still 1.
To get the full 5MHz, you need an amplitude of less than:
Vpk ≤ SR/(2*Π*5MHz)=0.063 V, that is a signal of less than 0.127 Vpp
I hope this helps.
Make sure you use the minimum SR in the datasheet.