The transformation you are describing is essentially a high-pass filter. Here is an intuitive explanation: suppose you feed a signal of magnitude 1 and of zero frequency in this filter. Then this signal is equal to $x[n]=1$. Hence the output of your filter will be $y[n]=0.05$. Observe the the energy of the output signal is much smaller than the energy of the input signal. Next, suppose that you feed a signal of small frequency in the filter. Because the frequency is small, adjacent samples will tend to be similar in magnitude and so the difference $x[n]-0.95 x[n-1]$ will tend to be very small. The numerical experiment that you are presenting, does not give you any information, because if $x[n]=0.34, x[n-1]=0.24$, then this does not say anything about the frequency of $x$. How about the other values of $x$, i.e. $x[n-2], x[n-3], e.t.c.$?