Since photon has no (rest)mass and
$$E^2=(pc)^2+(mc^2)^2$$
we derive that $E=pc$ for particle with no (rest)mass.
However, if we transform the non-relativistic formula for kinetic energy
$$E_k=\frac{mv^2}{2}$$ $$E_k=\frac{p^2}{2m}$$ $$E_k=\lim_{m\rightarrow0}\frac{p^2}{2m}=\lim_{m\rightarrow0}\frac{m^2v^2}{2m}=\lim_{m\rightarrow0}\frac{mv^2}{2}=\frac{pv}{2}$$
we derive that $E_k=\frac{pc}{2}$.
And this is bizarre because I'm expecting $E_k=E$.
I've just started learning physics recently so I'm pretty sure I've got certain part wrong. And here are my hypotheses:
1.$ $ The net energy of a photon $E_{net}=pc$ is equivalent $E_{net}=E_k+E_0=\frac{pc}{2}+E_0$ and thus $E_0=pc-E_k=\frac{pc}{2}$. (If that's true, what exactly is $E_0$? The rest energy of a photon?)
2.$ $ It is incorrect to derive the kinetic energy formula from integrating the equation
$$ \frac{\mathrm dE}{\mathrm dv}=mv $$
I should integrate another equation instead (perhaps $\frac{\mathrm dE}{\mathrm dv}=p$? If that's true, why do we break $p$ into $mv$ when deriving the formula $E_k=\frac{pc}{2}$?)
3.$ $ I got the limit wrong:
$$E_k=\lim_{m\rightarrow0}\frac{p^2}{2m}\ne \frac{pc}{2}$$
$$E_k=\lim_{m\rightarrow0}\frac{p^2}{2m}=pc$$
So is any of these hypothesis correct? And why?