How does the wavelength change in relativistic limit? In the text, it reads that the momentum of a particle will change if it is moving at speed close to light speed. In the general case, the wavelength is given as 
$$
  \lambda = \frac{h}{p}
$$
and 
$$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$
when $v \to c$, $p\to\infty$, so is it say that the wavelength is ZERO? I don't understand why the wavelength will change to zero if it is moving at speed very close to light speed?
 A: Lorentz contraction! The measured de Broglie wavelength in the direction of propagation vanishes because that's what special relativity says happens. The wavelength has to go as $h/p$ as you wrote, so why does it surprise you that when $p$ gets large the wavelength gets small?
A: Yes. The energy-momentum equation $E^2=p^2c^2+m^2c^4$ says that a massive object's mass (relativistic mass), momentum and energy approaches $\infty$ as a particle is accelerated towards $c$. There's nothing obvious about the fact that it requires infinite energy to accelerate it to $c$. This strictly means that you can't accelerate the object to speed-of-light...
Substituting Planck's wave equation $h\nu$  in $E=pc$ for photons, we get $$\lambda=\frac{hc}{pc}$$
Or in case of particles with rest energy where we can substitute for total energy, $$\lambda=\frac{h}{\sqrt{E^2-{(m_0c^2)}^2}}=\frac{h}{\sqrt{p^2-2mE_0}}$$
This equation says that matter waves are observable only for matter (massive) particles which always travel at $v<c$. In other words, there is no matter wave for such objects.
