= Solution
Work in the <shock frame>, assume a calorically <perfect gas> with fixed <specific-heat ratio> $\gamma>1$, and neglect magnetic stresses and body forces across the thin shock. The <internal energy> per unit mass is $e=p/[(\gamma-1)\rho]$, and the <specific enthalpy> is $w=\gamma p/[(\gamma-1)\rho]$. Integrating the local <conservation laws> across a thin stationary control volume makes the <mass flux> $\rho u$, <momentum flux> $p+\rho u^2$, and <energy flux> $u(\rho u^2/2+\gamma p/(\gamma-1))$ continuous. Hence the <Rankine-Hugoniot conditions for a perfect gas> are
$$
\rho_1u_1=\rho_2u_2=m,\qquad p_1+\rho_1u_1^2=p_2+\rho_2u_2^2,\qquad\frac{u_1^2}{2}+\frac{\gamma p_1}{(\gamma-1)\rho_1}=\frac{u_2^2}{2}+\frac{\gamma p_2}{(\gamma-1)\rho_2}.
$$
Multiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although <entropy production> occurs on the physical compression branch. The same constant in the <polytropic equation of state> $p=K\rho^\gamma$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.
In the <Strong-shock Rankine-Hugoniot conditions>, $p_1/(\rho_1u_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r=\rho_2/\rho_1$. The first two flux balances give $u_2=u_1/r$ and $p_2=\rho_1u_1^2(1-1/r)$ to leading order. The energy balance becomes
$$
\frac12=\frac1{2r^2}+\frac{\gamma}{\gamma-1}\frac{1-1/r}{r},\qquad(r-1)\{(\gamma-1)r-(\gamma+1)\}=0.
$$
Discarding the unchanged-state root $r=1$ gives the compressive <shock compression ratio> and the required limits:
$$
\boxed{\frac{\rho_2}{\rho_1}\longrightarrow\frac{\gamma+1}{\gamma-1},\qquad\frac{u_2}{u_1}\longrightarrow\frac{\gamma-1}{\gamma+1},\qquad\frac{p_2}{\rho_1u_1^2}\longrightarrow\frac2{\gamma+1}.}
$$
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