Note that:

$$ \lim_{S\rightarrow \infty} C(t,S) =S-K{\rm e}^{-r(T-t)} $$

for all $t$.

Basically because one can easily accept

$$ \lim_{S\rightarrow \infty} P(t,S) =0 $$

and one still expects the put-call parity to hold:

$$ C(t,S) - P(t,S) = S-K{\rm e}^{-r(T-t)} $$

for all $S$.

Note that:

$$ C(t,S) =S-K{\rm e}^{-r(T-t)} $$

as $S\rightarrow \infty$, for all $t$.

Basically because one can easily accept

$$ P(t,S) =0 $$

as $S\rightarrow \infty$, for all $t$,

and one still expects the put-call parity to hold:

$$ C(t,S) - P(t,S) = S-K{\rm e}^{-r(T-t)} $$

for all $S$.

Note that:

$$ \lim_{S\rightarrow \infty} C(t,S) =S-K{\rm e}^{-r(T-t)} $$

for all $t$.

Basically because one can easily accept

$$ \lim_{S\rightarrow \infty} P(t,S) =0 $$

and one still expects the put-call parity to hold:

$$ \lim_{S\rightarrow \infty} C(t,S) - \lim_{S\rightarrow \infty} P(t,S) =S-K{\rm e}^{-r(T-t)} $$

Note that:

$$ \lim_{S\rightarrow \infty} C(t,S) =S-K{\rm e}^{-r(T-t)} $$

for all $t$.

Basically because one can easily accept

$$ \lim_{S\rightarrow \infty} P(t,S) =0 $$

and one still expects the put-call parity to hold:

$$ C(t,S) - P(t,S) = S-K{\rm e}^{-r(T-t)} $$

for all $S$.