Ultra-Relativistic and Non-Relativistic cases for energy of a particle

For the non-relativistic case ($$mc^2 \gg pc$$) you have simplified too much.

A better approximation would be \begin{align} E &= (m^2c^4 + p^2c^2)^{1/2} \\ &= mc^2\left(1 + \frac{p^2c^2}{m^2c^4}\right)^{1/2} \\ &= mc^2\left(1 + \frac{p^2}{m^2c^2}\right)^{1/2} \\ &\approx mc^2\left(1 + \frac{p^2}{2m^2c^2}\right) \\ &= mc^2 + \frac{p^2}{2m}. \end{align}

Here you recognize $$\frac{p^2}{2m}$$ as the kinetic energy as known from Newtonian mechanics.

$$mc^2$$ is an additional constant (the rest energy) which causes no deviation from the Newtonian equations of motion.

This might be a useful energy-momentum diagram (below).

A particle's 4-momentum is drawn, as well its $$E$$ and $$p$$ components in this frame.

When describing limits, it's good to explicitly specify
(in addition to what is being varied) that which is being held constant.

So, if you keep the [rest] mass $$m$$ fixed, then increasing $$p$$ means that $$E$$ increases in such a way to keep $$E^2-(pc)^2=(mc^2)^2$$ fixed---that is, up along the "mass shell" (the hyperbola [or generally hyperboloid]).

It might be good to note that the components of the 4-momentum in this frame can be described by
$$E=mc^2\cosh\theta$$
and
$$pc=mc^2\sinh\theta$$
where $$\theta$$ is the Minkowski-angle (called the rapidity)
between the 4-momentum-vector and the vertical Energy axis.
So, $$(v/c)=\tanh\theta$$, $$\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\cosh\theta$$, and $$\gamma v=\frac{v}{\sqrt{1-(v/c)^2}}=\sinh\theta$$.

While keeping $$m$$ fixed, large $$p$$ corresponds to a large $$\theta$$.
Since $$\cosh\theta=\frac{1}{2}(e^\theta+e^{-\theta}$$)
and $$\sinh\theta=\frac{1}{2}(e^\theta-e^{-\theta}$$), as $$\theta$$ gets large, $$e^{-\theta}$$ gets small.
So, $$\cosh\theta$$ and $$\sinh\theta$$ approach each other---they approach $$e^\theta/2$$.
(In fact, it turns out that $$\cosh\theta-\sinh\theta=e^{-\theta}$$, and we said that $$e^{-\theta}$$ is getting small with larger $$\theta$$.)

[By the way, $$e^\theta=\cosh\theta+\sinh\theta=\cosh\theta(1+\tanh\theta)=\gamma(1+(v/c))=\sqrt{\frac{1+(v/c)}{1-(v/c)}}$$, the Doppler factor.]

[This image came from an old post of mine at Relativity and Momentum of photons ]