Xrfgtjtk

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

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

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

$$begin{align}
limlimits_{ato 0}A_a&=limlimits_{ato 0}dfrac{1}{a}intlimits_{-a/2}^{a/2}g(t)dttag{1},\ &=g(0)limlimits_{ato 0}dfrac{1}{a}intlimits_{-a/2}^{a/2}dttag{2}.
end{align}
$$

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:
$$begin{align}
limlimits_{ato 0}A_a&=g(0)limlimits_{ato 0}dfrac{1}{a}tag{3},
end{align}
$$
but this is wrong.

@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 $tto 0$.

Then,
$$
frac{1}{a}intlimits_{-a/2}^{a/2}g(t),dt=frac{1}{a}intlimits_{-a/2}^{a/2}[g(0)+h(t)],dt
$$
$$
=g(0)+frac{1}{a}intlimits_{-a/2}^{a/2}h(t),dt.
$$
Now, for any $epsilon>0$ it is possible to choose $a$ small enough such that $|h(t)|<epsilon$ for $xin [-a/2,a/2]$. For such $a$ we have
$$
|frac{1}{a}intlimits_{-a/2}^{a/2}h(t),dt|lefrac{1}{a}intlimits_{-a/2}^{a/2}epsilon,dt= epsilon
$$
thus implying that
$$
limlimits_{ato 0}frac{1}{a}|intlimits_{-a/2}^{a/2}h(t),dt|=0
$$
and you finally get
$$
limlimits_{ato 0}frac{1}{a}intlimits_{-a/2}^{a/2}g(t),dt=g(0)
$$
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.

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 _{ato 0}A_a=lim_{ato 0}frac{1}{a}int_{-a/2}^{a/2}g(t),dt\
&=lim_{ato 0}frac{1}{2a}int_{-a}^{a}g(t),dt\
&=lim_{ato 0}frac{G(a)-G(-a)}{2a}\
&=frac{1}{2}lim_{ato 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*}