= Solution
{c}
We give the explicit point $\boxed{x=1/4}$. Repeatedly choose the left remaining third and then the right remaining third. After $2m$ choices, the resulting closed <interval> is $[L_m,R_m]$, where
$$
L_m=\sum_{j=1}^m\frac2{3^{2j}}=\frac14(1-9^{-m}),\qquad
R_m=L_m+9^{-m}=\frac14+\frac34\,9^{-m}.
$$
These formulas follow from a finite <geometric series>. They show $L_m<1/4<R_m$, and that both endpoints tend to $1/4$. The point $1/4$ is also strictly inside each intermediate left third: its distance from $L_m$ is one quarter of the current length. It therefore never enters any assigned closed middle third or its endpoints, so $1/4\in C$.
Endpoint values in the <Cantor function> construction obey the same left/right choices, with halving in place of thirding. Thus
$$
f(L_m)=\sum_{j=1}^m\frac1{2^{2j}}=\frac13(1-4^{-m}),\qquad
f(R_m)=f(L_m)+4^{-m}=\frac13+\frac23\,4^{-m}.
$$
All these endpoints are in the domain of the partial <function>: $L_m$ is the right endpoint of a middle third assigned at round $2m$, and $R_m$ is the left endpoint of a middle third assigned at round $2m-1$.
Given $\varepsilon>0$, choose $m$ so that $(2/3)4^{-m}<\varepsilon$, and put $\delta=9^{-m}/4$. If $t\notin C$ and $|t-1/4|<\delta$, then $L_m<t<R_m$. The assumed <monotonicity> gives
$$
\frac13-\frac13\,4^{-m}\leq f(L_m)\leq f(t)\leq f(R_m)\leq\frac13+\frac23\,4^{-m}.
$$
Consequently $|f(t)-1/3|<\varepsilon$. This proves directly, on the stated partial domain,
$$
\boxed{\lim_{\substack{t\to1/4\\t\notin C}}f(t)=\frac13.}
$$
There are defined points arbitrarily close to $1/4$, for example $L_m$ and $R_m$, so the <limit of a function> is not vacuous. No <continuous extension> has been assumed. The residual $C$ here omits the endpoints that the question assigns values to; it is the usual <Cantor set> with those assigned endpoints removed.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/2021-mathematics-ii-cantor-limit.png]
{title=Assigned middle thirds and the nested intervals around $x=1/4$ that force the value $1/3$}
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