Is the limit of the mean value of a function around a point equal to the value of the function at that point? [closed]

It's true since \begin{align} |\frac{1}{2\delta}\int_{y - \delta}^{y + \delta}f(x)\,dx - f(y)| &= |\frac{1}{2\delta}\int_{y - \delta}^{y + \delta}f(x) - f(y)\,dx| \\ &\leq \frac{1}{2\delta}\int_{y - \delta}^{y + \delta}|f(x) - f(y)|\,dx \\ &\leq \frac{1}{2\delta}\sup_{x \in (y - \delta, y + \delta)}|f(x) - f(y)| \cdot 2\delta \\ &= \sup_{x \in (y - \delta, y + \delta)}|f(x) - f(y)| \to 0 \text{ as }\delta \to 0. \end{align} We know the last quantity goes to $0$ since that is the definition of continuity at $y$.

Yes, that is correct. By the integral mean value theorem, $$ \int\limits_{y-1/n}^{y+1/n}f(x)\,dx=2f(c)/n $$ where $c\in (y-1/n,y+1/n)$. As $n\to\infty$, $c\to y$ and, by continuity, $f(c)\to f(y)$.