= Solution
For $\rho\to1^+$, <Taylor expansion> gives $\rho^\gamma-1=\gamma(\rho-1)+O((\rho-1)^2)$. The <vanishing-cutoff limit of the Diesel cycle> is
$$
\boxed{\eta\longrightarrow1-\frac1{r^{\gamma-1}}
=1-\left(\frac{V_2}{V_1}\right)^{0.4}.}
$$
The maximum <temperature> tends to $T_2=T_1r^{\gamma-1}$, while the minimum stays at $T_1$. A Carnot engine between those limiting extrema has
$$
\eta_C=1-\frac{T_1}{T_2}=1-r^{1-\gamma},
$$
so \b[the limiting efficiencies coincide]. At exactly $V_3=V_2$, however, both <heat> input and net <work> vanish, and the cycle retraces its compression <adiabat>. The efficiency there is an undefined $0/0$; the boxed value is the continuous limit for neighbouring nonzero cycles, not finite <work> at zero <heat> input.
For a finite cutoff ratio $\rho>1$, the Carnot comparison uses $T_{\max}=T_3=T_1r^{\gamma-1}\rho$, giving $\eta_C=1-1/(r^{\gamma-1}\rho)$. The <Diesel cycle> has smaller efficiency. For example,
$$
\frac{\rho^\gamma-1}{\gamma(\rho-1)}
=\frac1{\rho-1}\int_1^\rho x^{\gamma-1}\,dx>1>\frac1\rho,
$$
which makes its rejected/input <heat> ratio larger than the Carnot ratio.
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