Op Amps: Bandwidth

The gain difference is that for non-inverting, Cf can only lower gain towards 1V/V; while for inverting Cf can reduce gain towards 0V/V.
I still confused 
why  non-inverting,Cf can only lower gain towards 1V/V and inverting Cf can reduce gain towards 0V/V this part could you explain in detail please

The explanation for that statement is that at high frequencies the capacitor can be considered a short. In the case of non-inverting configuration, if the capacitor is a short the op amp is in a buffer configuration, which has a gain of 1V/V. If the capacitor is a short in the inverting configuration, it bypasses the opamp in the signal path, thus giving a gain of 0V/V.

Mathematically where CF shorts out RF

Non Inverting gain is 1+RF/R1 so gain is 1 + 0/1k = 1

Inverting gain is -RF/R1 so gain is -0/1k = 0

I understand what's your mean, for non-inverting configuration at the high frequency the gain will be 20log(1V/V) =0dB ,

but for inverting configuration 20log(0V/V) is wrong value and the gain is -20dB ? 

Your mean is  20log(0V/V) is falling? 
but at high frequency there is a zero and pole right? I still don't understand this part 
I agree you said  real op amp drop with higher frequency   but how to explain 20log(0V/V)