How to explain this limit?

Question

Let $g(t)$ be a continuous function at $t=0$. Let $A_a$ be the following integral:

$$A_a = \dfrac{1}{a}\int\limits_{-a/2}^{a/2}g(t)dt.$$

Now, calculate the limit of $A_a$ as $a$ approaches $0$.

$$\begin{align} \lim\limits_{a\to 0}A_a&=\lim\limits_{a\to 0}\dfrac{1}{a}\int\limits_{-a/2}^{a/2}g(t)dt\tag{1},\\ &=g(0)\lim\limits_{a\to 0}\dfrac{1}{a}\int\limits_{-a/2}^{a/2}dt\tag{2}. \end{align} $$

I don't understand how can we go from $(1)$ to $(2)$. In fact, for me, as $a$ approaches $0$, the integral $\int_{-a/2}^{a/2}g(t)dt$ approahces $g(0)$. So the limit should be: $$\begin{align} \lim\limits_{a\to 0}A_a&=g(0)\lim\limits_{a\to 0}\dfrac{1}{a}\tag{3}, \end{align} $$ but this is wrong.

Answer 1

@zdm

One way to see how you go from $(1)$ to $(2)$ is the following. By continuity at $0$ you have: $$ g(t)=g(0)+h(t), $$ where $h(t)\to 0$ as $t\to 0$.

Then, $$ \frac{1}{a}\int\limits_{-a/2}^{a/2}g(t)\,dt=\frac{1}{a}\int\limits_{-a/2}^{a/2}[g(0)+h(t)]\,dt $$ $$ =g(0)+\frac{1}{a}\int\limits_{-a/2}^{a/2}h(t)\,dt. $$ Now, for any $\epsilon>0$ it is possible to choose $a$ small enough such that $|h(t)|<\epsilon$ for $x\in [-a/2,a/2]$. For such $a$ we have $$ |\frac{1}{a}\int\limits_{-a/2}^{a/2}h(t)\,dt|\le\frac{1}{a}\int\limits_{-a/2}^{a/2}\epsilon\,dt= \epsilon $$ thus implying that $$ \lim\limits_{a\to 0}\frac{1}{a}|\int\limits_{-a/2}^{a/2}h(t)\,dt|=0 $$ and you finally get $$ \lim\limits_{a\to 0}\frac{1}{a}\int\limits_{-a/2}^{a/2}g(t)\,dt=g(0) $$ Actually the idea behind is very simple: you are finding the average of a continuous function (at zero) on a sequence of inteervals shrinking to zero. Since the values stabilize around $g(0)$, the average does the same.

Hope this helps.

Answer 2

Best/simplest approach is to use Fundamental Theorem of Calculus. Let's assume that $g$ is Riemann integrable on some interval of type $[-h, h] $ otherwise your $A_a$ might not be defined.

Consider $$G(x) =\int_{0}^{x}g(t)\,dt$$ Since $g$ is continuous at $0$ it follows by Fundamental Theorem of Calculus that $G$ is differentiable at $0$ with $G'(0)=g(0)$.

Now we have \begin{align*} L&=\lim _{a\to 0}A_a=\lim_{a\to 0}\frac{1}{a}\int_{-a/2}^{a/2}g(t)\,dt\\ &=\lim_{a\to 0}\frac{1}{2a}\int_{-a}^{a}g(t)\,dt\\ &=\lim_{a\to 0}\frac{G(a)-G(-a)}{2a}\\ &=\frac{1}{2}\lim_{a\to 0}\left(\frac {G(a)-G(0)}{a}+\frac{G(-a)-G(0)}{-a} \right)\\ &=\frac{1}{2}(G'(0)+G'(0))\\ &=G'(0)=g(0) \end{align*}