How to explain this limit?












1












$begingroup$


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.






integration limits continuity







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Well first of all you need to say that your function is Riemann integrable on every $[c_1; c_2] subset [-a/2;a/2]$, otherwise your integrals are not defined. And no, you can’t just change the limit and integral: see, you have a $0/0$-type limit, try to use L'Hôpital's rule and when you will differentiate your integral in the numerator, you must use Leibniz integral rule for sure
    $endgroup$
    – Anton Zagrivin
    Jan 12 at 16:48


















1












$begingroup$


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.






integration limits continuity







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Well first of all you need to say that your function is Riemann integrable on every $[c_1; c_2] subset [-a/2;a/2]$, otherwise your integrals are not defined. And no, you can’t just change the limit and integral: see, you have a $0/0$-type limit, try to use L'Hôpital's rule and when you will differentiate your integral in the numerator, you must use Leibniz integral rule for sure
    $endgroup$
    – Anton Zagrivin
    Jan 12 at 16:48
















1












1








1





$begingroup$


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.






integration limits continuity







share|cite|improve this question









$endgroup$




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.






integration limits continuity




integration limits continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 16:29









zdmzdm

1876




1876








  • 1




    $begingroup$
    Well first of all you need to say that your function is Riemann integrable on every $[c_1; c_2] subset [-a/2;a/2]$, otherwise your integrals are not defined. And no, you can’t just change the limit and integral: see, you have a $0/0$-type limit, try to use L'Hôpital's rule and when you will differentiate your integral in the numerator, you must use Leibniz integral rule for sure
    $endgroup$
    – Anton Zagrivin
    Jan 12 at 16:48
















  • 1




    $begingroup$
    Well first of all you need to say that your function is Riemann integrable on every $[c_1; c_2] subset [-a/2;a/2]$, otherwise your integrals are not defined. And no, you can’t just change the limit and integral: see, you have a $0/0$-type limit, try to use L'Hôpital's rule and when you will differentiate your integral in the numerator, you must use Leibniz integral rule for sure
    $endgroup$
    – Anton Zagrivin
    Jan 12 at 16:48










1




1




$begingroup$
Well first of all you need to say that your function is Riemann integrable on every $[c_1; c_2] subset [-a/2;a/2]$, otherwise your integrals are not defined. And no, you can’t just change the limit and integral: see, you have a $0/0$-type limit, try to use L'Hôpital's rule and when you will differentiate your integral in the numerator, you must use Leibniz integral rule for sure
$endgroup$
– Anton Zagrivin
Jan 12 at 16:48






$begingroup$
Well first of all you need to say that your function is Riemann integrable on every $[c_1; c_2] subset [-a/2;a/2]$, otherwise your integrals are not defined. And no, you can’t just change the limit and integral: see, you have a $0/0$-type limit, try to use L'Hôpital's rule and when you will differentiate your integral in the numerator, you must use Leibniz integral rule for sure
$endgroup$
– Anton Zagrivin
Jan 12 at 16:48












2 Answers
2






active

oldest

votes


















2












$begingroup$

@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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The function $g$ (hence also $h$) should be integrable in a neighborhood of $0$. I'm not sure this is guaranteed only with the assumption that $g$ is continuous at $0$. Surely $g$ is bounded in a neighborhood of $0$, but this doesn't suffice.
    $endgroup$
    – egreg
    Jan 12 at 16:58










  • $begingroup$
    Agree. you have to assume that it Is integrable in some neighborhood of the origin (continuity on some (-delta, delta) would suffice of course.
    $endgroup$
    – GReyes
    Jan 12 at 18:04



















2












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great explanation! How do you go from $lim_{ato 0}frac{1}{a}int_{color{red}{-a/2}}^{color{red}{a/2}}g(t),dt$ to $ lim_{ato 0}frac{1}{color{red}{2a}}int_{color{red}{-a}}^{color{red}a}g(t),dt$?. P.S. To avoid having to write notag at the end of each line inside align environment you can consider use align* environment $ddotsmile$.
    $endgroup$
    – manooooh
    Jan 13 at 2:35








  • 1




    $begingroup$
    @manooooh: just replace $a$ by $2a$. Formally it is a substitution $a=2u$ (note that $uto 0$ as $ato 0$) and then replace the variable name $u$ by $a$ again. I will edit this to use align* and see how it looks.
    $endgroup$
    – Paramanand Singh
    Jan 13 at 5:19













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

@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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The function $g$ (hence also $h$) should be integrable in a neighborhood of $0$. I'm not sure this is guaranteed only with the assumption that $g$ is continuous at $0$. Surely $g$ is bounded in a neighborhood of $0$, but this doesn't suffice.
    $endgroup$
    – egreg
    Jan 12 at 16:58










  • $begingroup$
    Agree. you have to assume that it Is integrable in some neighborhood of the origin (continuity on some (-delta, delta) would suffice of course.
    $endgroup$
    – GReyes
    Jan 12 at 18:04
















2












$begingroup$

@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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The function $g$ (hence also $h$) should be integrable in a neighborhood of $0$. I'm not sure this is guaranteed only with the assumption that $g$ is continuous at $0$. Surely $g$ is bounded in a neighborhood of $0$, but this doesn't suffice.
    $endgroup$
    – egreg
    Jan 12 at 16:58










  • $begingroup$
    Agree. you have to assume that it Is integrable in some neighborhood of the origin (continuity on some (-delta, delta) would suffice of course.
    $endgroup$
    – GReyes
    Jan 12 at 18:04














2












2








2





$begingroup$

@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.






share|cite|improve this answer









$endgroup$



@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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 16:53









GReyesGReyes

1,99015




1,99015












  • $begingroup$
    The function $g$ (hence also $h$) should be integrable in a neighborhood of $0$. I'm not sure this is guaranteed only with the assumption that $g$ is continuous at $0$. Surely $g$ is bounded in a neighborhood of $0$, but this doesn't suffice.
    $endgroup$
    – egreg
    Jan 12 at 16:58










  • $begingroup$
    Agree. you have to assume that it Is integrable in some neighborhood of the origin (continuity on some (-delta, delta) would suffice of course.
    $endgroup$
    – GReyes
    Jan 12 at 18:04


















  • $begingroup$
    The function $g$ (hence also $h$) should be integrable in a neighborhood of $0$. I'm not sure this is guaranteed only with the assumption that $g$ is continuous at $0$. Surely $g$ is bounded in a neighborhood of $0$, but this doesn't suffice.
    $endgroup$
    – egreg
    Jan 12 at 16:58










  • $begingroup$
    Agree. you have to assume that it Is integrable in some neighborhood of the origin (continuity on some (-delta, delta) would suffice of course.
    $endgroup$
    – GReyes
    Jan 12 at 18:04
















$begingroup$
The function $g$ (hence also $h$) should be integrable in a neighborhood of $0$. I'm not sure this is guaranteed only with the assumption that $g$ is continuous at $0$. Surely $g$ is bounded in a neighborhood of $0$, but this doesn't suffice.
$endgroup$
– egreg
Jan 12 at 16:58




$begingroup$
The function $g$ (hence also $h$) should be integrable in a neighborhood of $0$. I'm not sure this is guaranteed only with the assumption that $g$ is continuous at $0$. Surely $g$ is bounded in a neighborhood of $0$, but this doesn't suffice.
$endgroup$
– egreg
Jan 12 at 16:58












$begingroup$
Agree. you have to assume that it Is integrable in some neighborhood of the origin (continuity on some (-delta, delta) would suffice of course.
$endgroup$
– GReyes
Jan 12 at 18:04




$begingroup$
Agree. you have to assume that it Is integrable in some neighborhood of the origin (continuity on some (-delta, delta) would suffice of course.
$endgroup$
– GReyes
Jan 12 at 18:04











2












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great explanation! How do you go from $lim_{ato 0}frac{1}{a}int_{color{red}{-a/2}}^{color{red}{a/2}}g(t),dt$ to $ lim_{ato 0}frac{1}{color{red}{2a}}int_{color{red}{-a}}^{color{red}a}g(t),dt$?. P.S. To avoid having to write notag at the end of each line inside align environment you can consider use align* environment $ddotsmile$.
    $endgroup$
    – manooooh
    Jan 13 at 2:35








  • 1




    $begingroup$
    @manooooh: just replace $a$ by $2a$. Formally it is a substitution $a=2u$ (note that $uto 0$ as $ato 0$) and then replace the variable name $u$ by $a$ again. I will edit this to use align* and see how it looks.
    $endgroup$
    – Paramanand Singh
    Jan 13 at 5:19


















2












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great explanation! How do you go from $lim_{ato 0}frac{1}{a}int_{color{red}{-a/2}}^{color{red}{a/2}}g(t),dt$ to $ lim_{ato 0}frac{1}{color{red}{2a}}int_{color{red}{-a}}^{color{red}a}g(t),dt$?. P.S. To avoid having to write notag at the end of each line inside align environment you can consider use align* environment $ddotsmile$.
    $endgroup$
    – manooooh
    Jan 13 at 2:35








  • 1




    $begingroup$
    @manooooh: just replace $a$ by $2a$. Formally it is a substitution $a=2u$ (note that $uto 0$ as $ato 0$) and then replace the variable name $u$ by $a$ again. I will edit this to use align* and see how it looks.
    $endgroup$
    – Paramanand Singh
    Jan 13 at 5:19
















2












2








2





$begingroup$

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*}






share|cite|improve this answer











$endgroup$



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*}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 at 5:20

























answered Jan 13 at 2:25









Paramanand SinghParamanand Singh

50.7k557168




50.7k557168












  • $begingroup$
    Great explanation! How do you go from $lim_{ato 0}frac{1}{a}int_{color{red}{-a/2}}^{color{red}{a/2}}g(t),dt$ to $ lim_{ato 0}frac{1}{color{red}{2a}}int_{color{red}{-a}}^{color{red}a}g(t),dt$?. P.S. To avoid having to write notag at the end of each line inside align environment you can consider use align* environment $ddotsmile$.
    $endgroup$
    – manooooh
    Jan 13 at 2:35








  • 1




    $begingroup$
    @manooooh: just replace $a$ by $2a$. Formally it is a substitution $a=2u$ (note that $uto 0$ as $ato 0$) and then replace the variable name $u$ by $a$ again. I will edit this to use align* and see how it looks.
    $endgroup$
    – Paramanand Singh
    Jan 13 at 5:19




















  • $begingroup$
    Great explanation! How do you go from $lim_{ato 0}frac{1}{a}int_{color{red}{-a/2}}^{color{red}{a/2}}g(t),dt$ to $ lim_{ato 0}frac{1}{color{red}{2a}}int_{color{red}{-a}}^{color{red}a}g(t),dt$?. P.S. To avoid having to write notag at the end of each line inside align environment you can consider use align* environment $ddotsmile$.
    $endgroup$
    – manooooh
    Jan 13 at 2:35








  • 1




    $begingroup$
    @manooooh: just replace $a$ by $2a$. Formally it is a substitution $a=2u$ (note that $uto 0$ as $ato 0$) and then replace the variable name $u$ by $a$ again. I will edit this to use align* and see how it looks.
    $endgroup$
    – Paramanand Singh
    Jan 13 at 5:19


















$begingroup$
Great explanation! How do you go from $lim_{ato 0}frac{1}{a}int_{color{red}{-a/2}}^{color{red}{a/2}}g(t),dt$ to $ lim_{ato 0}frac{1}{color{red}{2a}}int_{color{red}{-a}}^{color{red}a}g(t),dt$?. P.S. To avoid having to write notag at the end of each line inside align environment you can consider use align* environment $ddotsmile$.
$endgroup$
– manooooh
Jan 13 at 2:35






$begingroup$
Great explanation! How do you go from $lim_{ato 0}frac{1}{a}int_{color{red}{-a/2}}^{color{red}{a/2}}g(t),dt$ to $ lim_{ato 0}frac{1}{color{red}{2a}}int_{color{red}{-a}}^{color{red}a}g(t),dt$?. P.S. To avoid having to write notag at the end of each line inside align environment you can consider use align* environment $ddotsmile$.
$endgroup$
– manooooh
Jan 13 at 2:35






1




1




$begingroup$
@manooooh: just replace $a$ by $2a$. Formally it is a substitution $a=2u$ (note that $uto 0$ as $ato 0$) and then replace the variable name $u$ by $a$ again. I will edit this to use align* and see how it looks.
$endgroup$
– Paramanand Singh
Jan 13 at 5:19






$begingroup$
@manooooh: just replace $a$ by $2a$. Formally it is a substitution $a=2u$ (note that $uto 0$ as $ato 0$) and then replace the variable name $u$ by $a$ again. I will edit this to use align* and see how it looks.
$endgroup$
– Paramanand Singh
Jan 13 at 5:19




















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