 Solution

As $\sigma \downarrow 0$, the Black-Scholes formula tends to $\boxed{(S_0-K{e}^{-rT}{)}^{+}}$. This also follows by discounting the deterministic zero-volatility payoff; at equality $S_0=K{e}^{-rT}$ the limiting value is zero.

If $S_0>K{e}^{-rT}$ and a call trades at $C<S_0-K{e}^{-rT}$, buy a call, short one share, and buy a bond paying $K$ at maturity. Invest the positive initial surplus $S_0-K{e}^{-rT}-C$. At maturity the call, share and bond together pay $(S_T-K{)}^{+}-S_T+K=(K-S_T{)}^{+}\ge 0$, in addition to the strictly positive invested surplus. If the lower bound is zero, violating it means $C<0$: buy the call and invest the cash received, obtaining a nonnegative option payoff and a strictly positive bank payoff. Both strategies are arbitrages.