Suppose a function $q$ is continuous. Define $q^\ast(b)=\frac{1}{b-a}\int_a^b q(x)\;\mathrm{d}x$ to be the average of $q$ on the interval $[a,b]$.
Question: How do we prove that $\lim_{b\to a}q^\ast(b)=q(a)$? This is a very straightword intuitive result, but the proof has been eluding me.
I've done the work for when $q$ is monotonically increasing, which is shown below:
$\min_{x\in[a,b]}q(x)=q(a)$ and $\max_{x\in[a,b]}q(x)=q(b)$ {Property of monotonically increasing function}
If this is true, then we should be able to say that
$q(a)\leq\underbrace{\frac{1}{b-a}\int_a^b q(x)\;\mathrm{d}x}_{q^\ast}\leq q(b)$
The squeeze theorem can be used since $$\lim_{b\to a}q(b)=\lim_{b\to a}q(a)=q(a)$$
Thus, by the squeeze theorem, $$\lim_{b\to a}q^\ast(b)=q(a)$$
The argument for a decreasing function is essentially the same (with $q(a)$ in the place of $q(b)$).
Hint:
$$\min_{x \in [a, b]} q(x) \leq q^*(b) \leq \max_{x \in [a, b]} q(x).$$