Internet providers try to sell us the more and faster "fast internet" - 15Mb, 40Mb, 100Mb, and so on. But it's well known that the bandwidth the channel (in the most cases it's the telephone wire) restricted by the nature law. And this law is so fundamental as the light speed. (see, e.g., V.A.Kotetlnikov theorem). The bandwidth of a telephone wire is significantly less than providers propose us.
Can someone explain what is the matter here?
The propagation speed and the symbol rate are completely different concepts. The symbol rate is the rate at which you can transmit information (hence 'symbols'), and it is closely related to the bandwidth, which is the span of frequencies which can propagate well through the wire. The relationship between the symbol rate and the bandwidth was recently explored in
Why is bandwidth, range of frequencies, important when sending wave signals, such as in radio?
and my answer there goes a long way towards answering your question. Symbol rates of 100Mbps correspond to bandwidths of about 200MHz, which are perfectly reasonable for appropriate wires. Whether your ISP is actually implementing the technology to match their advertising - that's another matter, but the technology is certainly available.
The "fundamental law" you refer to is the Whittaker–Nyquist–Kotelnikov–Shannon theorem, and it simply states that if you want to observe a signal of frequency ~$\nu$, you should sample at about twice that frequency (I suggest you study the theorem statement closely). This is not really a problem for telephony - it simply dictates the rate at which you need to sample the signal. This sampling rate is roughly equal to the symbol rate, or slightly bigger by a small overhead. Sampling at 100Mbps is not a problem for a modern device, since processor speeds of >1GHz, ten times bigger, are prefectly possible.
