I connected the signal generator output to both the In+ and In-. Yet the waveform still have retained sine wave of very small value (needing 50000X gain). Why? won't it be rejected as well?
Since no amplifier has perfect common mode rejection, there will not be perfect cancellation. You will get some signal at the output, though it may be very small. The ratio between the AC amplitude of this output signal to the AC amplitude of the input signal is the common mode gain.
Note that you are lucky that the amplifier didn't saturate when you applied the same signal to both In+ and In-. All amplifiers have offset voltage. That is, In+ and In- need to be slightly different to make the output 0. That difference is the input offset voltage. If you multiply the input offset voltage by the gain of the amplifier, you get the output offset voltage. For some amplifiers, the open-loop output offset voltage is greater than the voltage swing available in the opamp. In your case, that is apparently not the case.
Im not measuring the CMRR. Just seeing if the same output going to In+ and In- of the [equipment] will get the signal cancelled. The picture shows it very clearly. But there is tiny sine wave retained..Why?
The common mode rejection ratio is the ratio of the common mode gain to the differential mode gain. Your test showed the common mode gain, so to find the common mode rejection ratio, you must also find the differential mode gain. There are a number of ways of doing this. For an open-loop op-amp, you must first bring your input signal down to a level that is sufficiently small that it does not saturate your op-amp. This can be achieved by using a voltage-divider. Then apply this attenuated signal to one input, and hold the other signal at some reference point. A good reference point might be the mid-point between the voltage rails.
Once you have found the common mode rejection ratio, you may want to convert it into decibels (dB). To do so, take the base-10 logarithm of the ratio, and multiply that number by 20.
I connected the signal generator output to both the In+ and In-.
is this the only thing you've done to measure the CMRR? Can you show a schematic instead? I don't think this is a proper method to measure CMRR. \$\endgroup\$