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How to compute $lim_x to 0 frac1x^2 int_0^x f(t)t space dt $?
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For a continuous function $f: R to R $ Define:
$$ lim_x to 0 frac1x^2 int_0^x f(t)t space dt $$
Since the function is continuous I can assume that it's also integrable since continuity implies integrability. I assume furthermore that there exists a function $F$ which is an antiderivative of $f$ for which the following are true:
$$int_a^b f(x) dx =F(b)-F(a) spacespacespacespacespace a,bin R space $$
$$ lim_x to x_0 fracF(x)-F(x_0)x-x_0=f(x)$$
And that for $f$
$$ lim_x to x_0 f(x)=f(x_0)$$
In order to find the limit, I used partial integration and ended up with:
$$ lim_x to 0 fracF(x)(x-1) +F(0)x^2$$
At this point, I tried to use L'Hôpital's rule and ended up with the value $fracf(0)2$ which seems totally wrong to me .
Any advice would be appreciated, I mainly think that my solution idea is wrong, but I am stuck.
limits derivatives definite-integrals continuity
edited yesterday
Asaf Karagila
291k31401731
asked 2 days ago
Konstantinos Zafeiris
1089
add a comment |Â
up vote
6
down vote
favorite
For a continuous function $f: R to R $ Define:
$$ lim_x to 0 frac1x^2 int_0^x f(t)t space dt $$
Since the function is continuous I can assume that it's also integrable since continuity implies integrability. I assume furthermore that there exists a function $F$ which is an antiderivative of $f$ for which the following are true:
$$int_a^b f(x) dx =F(b)-F(a) spacespacespacespacespace a,bin R space $$
$$ lim_x to x_0 fracF(x)-F(x_0)x-x_0=f(x)$$
And that for $f$
$$ lim_x to x_0 f(x)=f(x_0)$$
In order to find the limit, I used partial integration and ended up with:
$$ lim_x to 0 fracF(x)(x-1) +F(0)x^2$$
At this point, I tried to use L'Hôpital's rule and ended up with the value $fracf(0)2$ which seems totally wrong to me .
Any advice would be appreciated, I mainly think that my solution idea is wrong, but I am stuck.
limits derivatives definite-integrals continuity
edited yesterday
Asaf Karagila
291k31401731
asked 2 days ago
Konstantinos Zafeiris
1089
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
For a continuous function $f: R to R $ Define:
$$ lim_x to 0 frac1x^2 int_0^x f(t)t space dt $$
Since the function is continuous I can assume that it's also integrable since continuity implies integrability. I assume furthermore that there exists a function $F$ which is an antiderivative of $f$ for which the following are true:
$$int_a^b f(x) dx =F(b)-F(a) spacespacespacespacespace a,bin R space $$
$$ lim_x to x_0 fracF(x)-F(x_0)x-x_0=f(x)$$
And that for $f$
$$ lim_x to x_0 f(x)=f(x_0)$$
In order to find the limit, I used partial integration and ended up with:
$$ lim_x to 0 fracF(x)(x-1) +F(0)x^2$$
At this point, I tried to use L'Hôpital's rule and ended up with the value $fracf(0)2$ which seems totally wrong to me .
Any advice would be appreciated, I mainly think that my solution idea is wrong, but I am stuck.
limits derivatives definite-integrals continuity
edited yesterday
Asaf Karagila
291k31401731
asked 2 days ago
Konstantinos Zafeiris
1089
For a continuous function $f: R to R $ Define:
$$ lim_x to 0 frac1x^2 int_0^x f(t)t space dt $$
Since the function is continuous I can assume that it's also integrable since continuity implies integrability. I assume furthermore that there exists a function $F$ which is an antiderivative of $f$ for which the following are true:
$$int_a^b f(x) dx =F(b)-F(a) spacespacespacespacespace a,bin R space $$
$$ lim_x to x_0 fracF(x)-F(x_0)x-x_0=f(x)$$
And that for $f$
$$ lim_x to x_0 f(x)=f(x_0)$$
In order to find the limit, I used partial integration and ended up with:
$$ lim_x to 0 fracF(x)(x-1) +F(0)x^2$$
At this point, I tried to use L'Hôpital's rule and ended up with the value $fracf(0)2$ which seems totally wrong to me .
Any advice would be appreciated, I mainly think that my solution idea is wrong, but I am stuck.
limits derivatives definite-integrals continuity
edited yesterday
Asaf Karagila
291k31401731
asked 2 days ago
Konstantinos Zafeiris
1089
edited yesterday
Asaf Karagila
291k31401731
edited yesterday
Asaf Karagila
291k31401731
edited yesterday
Asaf Karagila
291k31401731
291k31401731
asked 2 days ago
Konstantinos Zafeiris
1089
asked 2 days ago
Konstantinos Zafeiris
1089
asked 2 days ago
Konstantinos Zafeiris
1089
1089
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
6
down vote
accepted
By L'Hôpital's rule, this limit is equal to
$$lim_xto 0 fracxf(x)2x = frac12lim_xto 0 f(x) = fracf(0)2.$$
(Use the first Fundamental Theorem of Calculus to differentiate the integral, since the integrand is guaranteed continuous.) You were correct.
answered 2 days ago
Ted Shifrin
59.4k44386
Even though I've found the correct answer, I still feel that I've taken a shortcut or a path that's not permitted. Is the limit (which I've stated) after I used partial integration correct or should I have handled the integration in another way in order to reach $ lim_xto0fracx,f(x)2x $ ?
– Konstantinos Zafeiris
2 days ago
You don't need to integrate. Use the part of the FTC that tells you how to differentiate $int_a^x g(t),dt$ when $g$ is continuous. I don't follow your solution. If you're going to use integration by parts, you'll need to integrate $int_0^x F(t),dt$, won't you?
– Ted Shifrin
2 days ago
Ok, now I see, thank you.
– Konstantinos Zafeiris
2 days ago
add a comment |Â
up vote
5
down vote
The limit is indeed $f(0)/2$.
L'Hôpital's Rule applies because the limit of the quotient of the derivatives of numerator and denominator exists. Thus
$$
lim_xto0frac1x^2int_0^x f(t),t,dt
=lim_xto0fracx,f(x)2x=fracf(0)2
$$
answered 2 days ago
Martin Argerami
115k1071164
But if you use the antiderivative you get: $$lim_x to 0 frac1x^2 int_0^x f(t)t space dt = lim_x to 0 frac1x^2 (F(x)-F(0)) =\ lim_x to 0frac 1 x fracF(x)-F(0)x =bigg(lim_x to 0frac 1 xbigg)bigg( lim_x to 0 fracF(x)-F(0)xbigg) =\ pm infty cdot f(x)$$
– Sudix
2 days ago
@Sudix You get $lim_xto0fracF(x)-F(0)x^2$. You can only separate the limit as you did when the limit of the factors exist and some extra cases, but not in the case that one for the factors tends to $infty$ and the other to $0=f(0)cdot 0$ (not $f(x )$ as you wrote) as happens here.
– spiralstotheleft
2 days ago
@spiralstotheleft You're right, I forgot about the case $f(0)=0$, but if $f(0) not = 0$, we can do this seperation, so something still is off
– Sudix
2 days ago
@Sudix The factor $lim_xto0fracF(x)-F(0)x$ is $(f(x)cdot x)|_x=0$, if $f(0)neq0$ the factor is still zero.
– spiralstotheleft
2 days ago
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– spiralstotheleft
2 days ago
 |Â
show 2 more comments
up vote
4
down vote
An approach not relying on L'Hopital's Rule.
We have, for $xneq 0$ and with the change of variable $u=fractx$, $$
frac1x^2int_0^x t f(t)dt
= int_0^1 u f(xu) du
$$
Now, it is not hard to prove that
$$
lim_xto 0int_0^1 u f(xu) du
= int_0^1 u lim_xto 0 f(xu) du
= int_0^1 u f(0) du
= f(0)left [fracx^22right]^1_0
= fracf(0)2
$$
(where the swapping limit/integral can be justified e.g. by arguing about uniform convergence, using continuity of $f$).
answered 2 days ago
Clement C.
46.9k33682
2
Nice approach. You don't even need to think about things like uniform convergence if you instead consider $left|int_0^1 uf(xu),du - int_0^1 uf(0),duright|le int_0^1 u|f(xu)-f(0)|du< epsilonint_0^1 u,du$ whenever $x<delta$ by continuity (since $xule x$ for all $0le ule 1$).
– Ted Shifrin
yesterday
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
yesterday
@TedShifrin Yes, good point.
– Clement C.
yesterday
add a comment |Â
up vote
0
down vote
Write $f(t) $ as $f(t) - f(0)+f(0)$ and then the desired limit is $$lim_xto 0frac1x^2int_0^xf(t)-f(0)t,dt+fracf(0)2tag1$$ Next we can show that the first limit above is $0$. Let $epsilon >0$ be given. Then by continuity we have a $delta>0$ such that $$|f(x) - f(0)|<epsilon $$ whenever $|x|<delta$. Thus we have $$left|int_0^xf(t)-f(0)t,dtright |leqint_0^x|f(t)-f(0)|t,dt<fracepsilon x^22$$ whenever $0<x<delta$. Similar inequality holds when $-delta <x<0$. It thus follows that the first limit in $(1)$ above is $0$. Thus the desired limit is $f(0)/2$.
answered yesterday
Paramanand Singh
45k553142
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
By L'Hôpital's rule, this limit is equal to
$$lim_xto 0 fracxf(x)2x = frac12lim_xto 0 f(x) = fracf(0)2.$$
(Use the first Fundamental Theorem of Calculus to differentiate the integral, since the integrand is guaranteed continuous.) You were correct.
answered 2 days ago
Ted Shifrin
59.4k44386
Even though I've found the correct answer, I still feel that I've taken a shortcut or a path that's not permitted. Is the limit (which I've stated) after I used partial integration correct or should I have handled the integration in another way in order to reach $ lim_xto0fracx,f(x)2x $ ?
– Konstantinos Zafeiris
2 days ago
You don't need to integrate. Use the part of the FTC that tells you how to differentiate $int_a^x g(t),dt$ when $g$ is continuous. I don't follow your solution. If you're going to use integration by parts, you'll need to integrate $int_0^x F(t),dt$, won't you?
– Ted Shifrin
2 days ago
Ok, now I see, thank you.
– Konstantinos Zafeiris
2 days ago
add a comment |Â
up vote
6
down vote
accepted
By L'Hôpital's rule, this limit is equal to
$$lim_xto 0 fracxf(x)2x = frac12lim_xto 0 f(x) = fracf(0)2.$$
(Use the first Fundamental Theorem of Calculus to differentiate the integral, since the integrand is guaranteed continuous.) You were correct.
answered 2 days ago
Ted Shifrin
59.4k44386
Even though I've found the correct answer, I still feel that I've taken a shortcut or a path that's not permitted. Is the limit (which I've stated) after I used partial integration correct or should I have handled the integration in another way in order to reach $ lim_xto0fracx,f(x)2x $ ?
– Konstantinos Zafeiris
2 days ago
You don't need to integrate. Use the part of the FTC that tells you how to differentiate $int_a^x g(t),dt$ when $g$ is continuous. I don't follow your solution. If you're going to use integration by parts, you'll need to integrate $int_0^x F(t),dt$, won't you?
– Ted Shifrin
2 days ago
Ok, now I see, thank you.
– Konstantinos Zafeiris
2 days ago
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
By L'Hôpital's rule, this limit is equal to
$$lim_xto 0 fracxf(x)2x = frac12lim_xto 0 f(x) = fracf(0)2.$$
(Use the first Fundamental Theorem of Calculus to differentiate the integral, since the integrand is guaranteed continuous.) You were correct.
answered 2 days ago
Ted Shifrin
59.4k44386
By L'Hôpital's rule, this limit is equal to
$$lim_xto 0 fracxf(x)2x = frac12lim_xto 0 f(x) = fracf(0)2.$$
(Use the first Fundamental Theorem of Calculus to differentiate the integral, since the integrand is guaranteed continuous.) You were correct.
answered 2 days ago
Ted Shifrin
59.4k44386
answered 2 days ago
Ted Shifrin
59.4k44386
answered 2 days ago
Ted Shifrin
59.4k44386
answered 2 days ago
Ted Shifrin
59.4k44386
59.4k44386
Even though I've found the correct answer, I still feel that I've taken a shortcut or a path that's not permitted. Is the limit (which I've stated) after I used partial integration correct or should I have handled the integration in another way in order to reach $ lim_xto0fracx,f(x)2x $ ?
– Konstantinos Zafeiris
2 days ago
You don't need to integrate. Use the part of the FTC that tells you how to differentiate $int_a^x g(t),dt$ when $g$ is continuous. I don't follow your solution. If you're going to use integration by parts, you'll need to integrate $int_0^x F(t),dt$, won't you?
– Ted Shifrin
2 days ago
Ok, now I see, thank you.
– Konstantinos Zafeiris
2 days ago
add a comment |Â
Even though I've found the correct answer, I still feel that I've taken a shortcut or a path that's not permitted. Is the limit (which I've stated) after I used partial integration correct or should I have handled the integration in another way in order to reach $ lim_xto0fracx,f(x)2x $ ?
– Konstantinos Zafeiris
2 days ago
You don't need to integrate. Use the part of the FTC that tells you how to differentiate $int_a^x g(t),dt$ when $g$ is continuous. I don't follow your solution. If you're going to use integration by parts, you'll need to integrate $int_0^x F(t),dt$, won't you?
– Ted Shifrin
2 days ago
Ok, now I see, thank you.
– Konstantinos Zafeiris
2 days ago
Even though I've found the correct answer, I still feel that I've taken a shortcut or a path that's not permitted. Is the limit (which I've stated) after I used partial integration correct or should I have handled the integration in another way in order to reach $ lim_xto0fracx,f(x)2x $ ?
– Konstantinos Zafeiris
2 days ago
Even though I've found the correct answer, I still feel that I've taken a shortcut or a path that's not permitted. Is the limit (which I've stated) after I used partial integration correct or should I have handled the integration in another way in order to reach $ lim_xto0fracx,f(x)2x $ ?
– Konstantinos Zafeiris
2 days ago
You don't need to integrate. Use the part of the FTC that tells you how to differentiate $int_a^x g(t),dt$ when $g$ is continuous. I don't follow your solution. If you're going to use integration by parts, you'll need to integrate $int_0^x F(t),dt$, won't you?
– Ted Shifrin
2 days ago
You don't need to integrate. Use the part of the FTC that tells you how to differentiate $int_a^x g(t),dt$ when $g$ is continuous. I don't follow your solution. If you're going to use integration by parts, you'll need to integrate $int_0^x F(t),dt$, won't you?
– Ted Shifrin
2 days ago
Ok, now I see, thank you.
– Konstantinos Zafeiris
2 days ago
Ok, now I see, thank you.
– Konstantinos Zafeiris
2 days ago
add a comment |Â
up vote
5
down vote
The limit is indeed $f(0)/2$.
L'Hôpital's Rule applies because the limit of the quotient of the derivatives of numerator and denominator exists. Thus
$$
lim_xto0frac1x^2int_0^x f(t),t,dt
=lim_xto0fracx,f(x)2x=fracf(0)2
$$
answered 2 days ago
Martin Argerami
115k1071164
But if you use the antiderivative you get: $$lim_x to 0 frac1x^2 int_0^x f(t)t space dt = lim_x to 0 frac1x^2 (F(x)-F(0)) =\ lim_x to 0frac 1 x fracF(x)-F(0)x =bigg(lim_x to 0frac 1 xbigg)bigg( lim_x to 0 fracF(x)-F(0)xbigg) =\ pm infty cdot f(x)$$
– Sudix
2 days ago
@Sudix You get $lim_xto0fracF(x)-F(0)x^2$. You can only separate the limit as you did when the limit of the factors exist and some extra cases, but not in the case that one for the factors tends to $infty$ and the other to $0=f(0)cdot 0$ (not $f(x )$ as you wrote) as happens here.
– spiralstotheleft
2 days ago
@spiralstotheleft You're right, I forgot about the case $f(0)=0$, but if $f(0) not = 0$, we can do this seperation, so something still is off
– Sudix
2 days ago
@Sudix The factor $lim_xto0fracF(x)-F(0)x$ is $(f(x)cdot x)|_x=0$, if $f(0)neq0$ the factor is still zero.
– spiralstotheleft
2 days ago
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– spiralstotheleft
2 days ago
 |Â
show 2 more comments
up vote
5
down vote
The limit is indeed $f(0)/2$.
L'Hôpital's Rule applies because the limit of the quotient of the derivatives of numerator and denominator exists. Thus
$$
lim_xto0frac1x^2int_0^x f(t),t,dt
=lim_xto0fracx,f(x)2x=fracf(0)2
$$
answered 2 days ago
Martin Argerami
115k1071164
But if you use the antiderivative you get: $$lim_x to 0 frac1x^2 int_0^x f(t)t space dt = lim_x to 0 frac1x^2 (F(x)-F(0)) =\ lim_x to 0frac 1 x fracF(x)-F(0)x =bigg(lim_x to 0frac 1 xbigg)bigg( lim_x to 0 fracF(x)-F(0)xbigg) =\ pm infty cdot f(x)$$
– Sudix
2 days ago
@Sudix You get $lim_xto0fracF(x)-F(0)x^2$. You can only separate the limit as you did when the limit of the factors exist and some extra cases, but not in the case that one for the factors tends to $infty$ and the other to $0=f(0)cdot 0$ (not $f(x )$ as you wrote) as happens here.
– spiralstotheleft
2 days ago
@spiralstotheleft You're right, I forgot about the case $f(0)=0$, but if $f(0) not = 0$, we can do this seperation, so something still is off
– Sudix
2 days ago
@Sudix The factor $lim_xto0fracF(x)-F(0)x$ is $(f(x)cdot x)|_x=0$, if $f(0)neq0$ the factor is still zero.
– spiralstotheleft
2 days ago
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– spiralstotheleft
2 days ago
 |Â
show 2 more comments
up vote
5
down vote
up vote
5
down vote
The limit is indeed $f(0)/2$.
L'Hôpital's Rule applies because the limit of the quotient of the derivatives of numerator and denominator exists. Thus
$$
lim_xto0frac1x^2int_0^x f(t),t,dt
=lim_xto0fracx,f(x)2x=fracf(0)2
$$
answered 2 days ago
Martin Argerami
115k1071164
The limit is indeed $f(0)/2$.
L'Hôpital's Rule applies because the limit of the quotient of the derivatives of numerator and denominator exists. Thus
$$
lim_xto0frac1x^2int_0^x f(t),t,dt
=lim_xto0fracx,f(x)2x=fracf(0)2
$$
answered 2 days ago
Martin Argerami
115k1071164
answered 2 days ago
Martin Argerami
115k1071164
answered 2 days ago
Martin Argerami
115k1071164
answered 2 days ago
Martin Argerami
115k1071164
115k1071164
But if you use the antiderivative you get: $$lim_x to 0 frac1x^2 int_0^x f(t)t space dt = lim_x to 0 frac1x^2 (F(x)-F(0)) =\ lim_x to 0frac 1 x fracF(x)-F(0)x =bigg(lim_x to 0frac 1 xbigg)bigg( lim_x to 0 fracF(x)-F(0)xbigg) =\ pm infty cdot f(x)$$
– Sudix
2 days ago
@Sudix You get $lim_xto0fracF(x)-F(0)x^2$. You can only separate the limit as you did when the limit of the factors exist and some extra cases, but not in the case that one for the factors tends to $infty$ and the other to $0=f(0)cdot 0$ (not $f(x )$ as you wrote) as happens here.
– spiralstotheleft
2 days ago
@spiralstotheleft You're right, I forgot about the case $f(0)=0$, but if $f(0) not = 0$, we can do this seperation, so something still is off
– Sudix
2 days ago
@Sudix The factor $lim_xto0fracF(x)-F(0)x$ is $(f(x)cdot x)|_x=0$, if $f(0)neq0$ the factor is still zero.
– spiralstotheleft
2 days ago
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– spiralstotheleft
2 days ago
 |Â
show 2 more comments
But if you use the antiderivative you get: $$lim_x to 0 frac1x^2 int_0^x f(t)t space dt = lim_x to 0 frac1x^2 (F(x)-F(0)) =\ lim_x to 0frac 1 x fracF(x)-F(0)x =bigg(lim_x to 0frac 1 xbigg)bigg( lim_x to 0 fracF(x)-F(0)xbigg) =\ pm infty cdot f(x)$$
– Sudix
2 days ago
@Sudix You get $lim_xto0fracF(x)-F(0)x^2$. You can only separate the limit as you did when the limit of the factors exist and some extra cases, but not in the case that one for the factors tends to $infty$ and the other to $0=f(0)cdot 0$ (not $f(x )$ as you wrote) as happens here.
– spiralstotheleft
2 days ago
@spiralstotheleft You're right, I forgot about the case $f(0)=0$, but if $f(0) not = 0$, we can do this seperation, so something still is off
– Sudix
2 days ago
@Sudix The factor $lim_xto0fracF(x)-F(0)x$ is $(f(x)cdot x)|_x=0$, if $f(0)neq0$ the factor is still zero.
– spiralstotheleft
2 days ago
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– spiralstotheleft
2 days ago
But if you use the antiderivative you get: $$lim_x to 0 frac1x^2 int_0^x f(t)t space dt = lim_x to 0 frac1x^2 (F(x)-F(0)) =\ lim_x to 0frac 1 x fracF(x)-F(0)x =bigg(lim_x to 0frac 1 xbigg)bigg( lim_x to 0 fracF(x)-F(0)xbigg) =\ pm infty cdot f(x)$$
– Sudix
2 days ago
But if you use the antiderivative you get: $$lim_x to 0 frac1x^2 int_0^x f(t)t space dt = lim_x to 0 frac1x^2 (F(x)-F(0)) =\ lim_x to 0frac 1 x fracF(x)-F(0)x =bigg(lim_x to 0frac 1 xbigg)bigg( lim_x to 0 fracF(x)-F(0)xbigg) =\ pm infty cdot f(x)$$
– Sudix
2 days ago
@Sudix You get $lim_xto0fracF(x)-F(0)x^2$. You can only separate the limit as you did when the limit of the factors exist and some extra cases, but not in the case that one for the factors tends to $infty$ and the other to $0=f(0)cdot 0$ (not $f(x )$ as you wrote) as happens here.
– spiralstotheleft
2 days ago
@Sudix You get $lim_xto0fracF(x)-F(0)x^2$. You can only separate the limit as you did when the limit of the factors exist and some extra cases, but not in the case that one for the factors tends to $infty$ and the other to $0=f(0)cdot 0$ (not $f(x )$ as you wrote) as happens here.
– spiralstotheleft
2 days ago
@spiralstotheleft You're right, I forgot about the case $f(0)=0$, but if $f(0) not = 0$, we can do this seperation, so something still is off
– Sudix
2 days ago
@spiralstotheleft You're right, I forgot about the case $f(0)=0$, but if $f(0) not = 0$, we can do this seperation, so something still is off
– Sudix
2 days ago
@Sudix The factor $lim_xto0fracF(x)-F(0)x$ is $(f(x)cdot x)|_x=0$, if $f(0)neq0$ the factor is still zero.
– spiralstotheleft
2 days ago
@Sudix The factor $lim_xto0fracF(x)-F(0)x$ is $(f(x)cdot x)|_x=0$, if $f(0)neq0$ the factor is still zero.
– spiralstotheleft
2 days ago
1
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– spiralstotheleft
2 days ago
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– spiralstotheleft
2 days ago
 |Â
show 2 more comments
up vote
4
down vote
An approach not relying on L'Hopital's Rule.
We have, for $xneq 0$ and with the change of variable $u=fractx$, $$
frac1x^2int_0^x t f(t)dt
= int_0^1 u f(xu) du
$$
Now, it is not hard to prove that
$$
lim_xto 0int_0^1 u f(xu) du
= int_0^1 u lim_xto 0 f(xu) du
= int_0^1 u f(0) du
= f(0)left [fracx^22right]^1_0
= fracf(0)2
$$
(where the swapping limit/integral can be justified e.g. by arguing about uniform convergence, using continuity of $f$).
answered 2 days ago
Clement C.
46.9k33682
2
Nice approach. You don't even need to think about things like uniform convergence if you instead consider $left|int_0^1 uf(xu),du - int_0^1 uf(0),duright|le int_0^1 u|f(xu)-f(0)|du< epsilonint_0^1 u,du$ whenever $x<delta$ by continuity (since $xule x$ for all $0le ule 1$).
– Ted Shifrin
yesterday
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
yesterday
@TedShifrin Yes, good point.
– Clement C.
yesterday
add a comment |Â
up vote
4
down vote
An approach not relying on L'Hopital's Rule.
We have, for $xneq 0$ and with the change of variable $u=fractx$, $$
frac1x^2int_0^x t f(t)dt
= int_0^1 u f(xu) du
$$
Now, it is not hard to prove that
$$
lim_xto 0int_0^1 u f(xu) du
= int_0^1 u lim_xto 0 f(xu) du
= int_0^1 u f(0) du
= f(0)left [fracx^22right]^1_0
= fracf(0)2
$$
(where the swapping limit/integral can be justified e.g. by arguing about uniform convergence, using continuity of $f$).
answered 2 days ago
Clement C.
46.9k33682
2
Nice approach. You don't even need to think about things like uniform convergence if you instead consider $left|int_0^1 uf(xu),du - int_0^1 uf(0),duright|le int_0^1 u|f(xu)-f(0)|du< epsilonint_0^1 u,du$ whenever $x<delta$ by continuity (since $xule x$ for all $0le ule 1$).
– Ted Shifrin
yesterday
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
yesterday
@TedShifrin Yes, good point.
– Clement C.
yesterday
add a comment |Â
up vote
4
down vote
up vote
4
down vote
An approach not relying on L'Hopital's Rule.
We have, for $xneq 0$ and with the change of variable $u=fractx$, $$
frac1x^2int_0^x t f(t)dt
= int_0^1 u f(xu) du
$$
Now, it is not hard to prove that
$$
lim_xto 0int_0^1 u f(xu) du
= int_0^1 u lim_xto 0 f(xu) du
= int_0^1 u f(0) du
= f(0)left [fracx^22right]^1_0
= fracf(0)2
$$
(where the swapping limit/integral can be justified e.g. by arguing about uniform convergence, using continuity of $f$).
answered 2 days ago
Clement C.
46.9k33682
An approach not relying on L'Hopital's Rule.
We have, for $xneq 0$ and with the change of variable $u=fractx$, $$
frac1x^2int_0^x t f(t)dt
= int_0^1 u f(xu) du
$$
Now, it is not hard to prove that
$$
lim_xto 0int_0^1 u f(xu) du
= int_0^1 u lim_xto 0 f(xu) du
= int_0^1 u f(0) du
= f(0)left [fracx^22right]^1_0
= fracf(0)2
$$
(where the swapping limit/integral can be justified e.g. by arguing about uniform convergence, using continuity of $f$).
answered 2 days ago
Clement C.
46.9k33682
edited 2 days ago
answered 2 days ago
Clement C.
46.9k33682
answered 2 days ago
Clement C.
46.9k33682
answered 2 days ago
Clement C.
46.9k33682
46.9k33682
2
Nice approach. You don't even need to think about things like uniform convergence if you instead consider $left|int_0^1 uf(xu),du - int_0^1 uf(0),duright|le int_0^1 u|f(xu)-f(0)|du< epsilonint_0^1 u,du$ whenever $x<delta$ by continuity (since $xule x$ for all $0le ule 1$).
– Ted Shifrin
yesterday
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
yesterday
@TedShifrin Yes, good point.
– Clement C.
yesterday
add a comment |Â
2
Nice approach. You don't even need to think about things like uniform convergence if you instead consider $left|int_0^1 uf(xu),du - int_0^1 uf(0),duright|le int_0^1 u|f(xu)-f(0)|du< epsilonint_0^1 u,du$ whenever $x<delta$ by continuity (since $xule x$ for all $0le ule 1$).
– Ted Shifrin
yesterday
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
yesterday
@TedShifrin Yes, good point.
– Clement C.
yesterday
2
2
Nice approach. You don't even need to think about things like uniform convergence if you instead consider $left|int_0^1 uf(xu),du - int_0^1 uf(0),duright|le int_0^1 u|f(xu)-f(0)|du< epsilonint_0^1 u,du$ whenever $x<delta$ by continuity (since $xule x$ for all $0le ule 1$).
– Ted Shifrin
yesterday
Nice approach. You don't even need to think about things like uniform convergence if you instead consider $left|int_0^1 uf(xu),du - int_0^1 uf(0),duright|le int_0^1 u|f(xu)-f(0)|du< epsilonint_0^1 u,du$ whenever $x<delta$ by continuity (since $xule x$ for all $0le ule 1$).
– Ted Shifrin
yesterday
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
yesterday
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
yesterday
@TedShifrin Yes, good point.
– Clement C.
yesterday
@TedShifrin Yes, good point.
– Clement C.
yesterday
add a comment |Â
up vote
0
down vote
Write $f(t) $ as $f(t) - f(0)+f(0)$ and then the desired limit is $$lim_xto 0frac1x^2int_0^xf(t)-f(0)t,dt+fracf(0)2tag1$$ Next we can show that the first limit above is $0$. Let $epsilon >0$ be given. Then by continuity we have a $delta>0$ such that $$|f(x) - f(0)|<epsilon $$ whenever $|x|<delta$. Thus we have $$left|int_0^xf(t)-f(0)t,dtright |leqint_0^x|f(t)-f(0)|t,dt<fracepsilon x^22$$ whenever $0<x<delta$. Similar inequality holds when $-delta <x<0$. It thus follows that the first limit in $(1)$ above is $0$. Thus the desired limit is $f(0)/2$.
answered yesterday
Paramanand Singh
45k553142
add a comment |Â
up vote
0
down vote
Write $f(t) $ as $f(t) - f(0)+f(0)$ and then the desired limit is $$lim_xto 0frac1x^2int_0^xf(t)-f(0)t,dt+fracf(0)2tag1$$ Next we can show that the first limit above is $0$. Let $epsilon >0$ be given. Then by continuity we have a $delta>0$ such that $$|f(x) - f(0)|<epsilon $$ whenever $|x|<delta$. Thus we have $$left|int_0^xf(t)-f(0)t,dtright |leqint_0^x|f(t)-f(0)|t,dt<fracepsilon x^22$$ whenever $0<x<delta$. Similar inequality holds when $-delta <x<0$. It thus follows that the first limit in $(1)$ above is $0$. Thus the desired limit is $f(0)/2$.
answered yesterday
Paramanand Singh
45k553142
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Write $f(t) $ as $f(t) - f(0)+f(0)$ and then the desired limit is $$lim_xto 0frac1x^2int_0^xf(t)-f(0)t,dt+fracf(0)2tag1$$ Next we can show that the first limit above is $0$. Let $epsilon >0$ be given. Then by continuity we have a $delta>0$ such that $$|f(x) - f(0)|<epsilon $$ whenever $|x|<delta$. Thus we have $$left|int_0^xf(t)-f(0)t,dtright |leqint_0^x|f(t)-f(0)|t,dt<fracepsilon x^22$$ whenever $0<x<delta$. Similar inequality holds when $-delta <x<0$. It thus follows that the first limit in $(1)$ above is $0$. Thus the desired limit is $f(0)/2$.
answered yesterday
Paramanand Singh
45k553142
Write $f(t) $ as $f(t) - f(0)+f(0)$ and then the desired limit is $$lim_xto 0frac1x^2int_0^xf(t)-f(0)t,dt+fracf(0)2tag1$$ Next we can show that the first limit above is $0$. Let $epsilon >0$ be given. Then by continuity we have a $delta>0$ such that $$|f(x) - f(0)|<epsilon $$ whenever $|x|<delta$. Thus we have $$left|int_0^xf(t)-f(0)t,dtright |leqint_0^x|f(t)-f(0)|t,dt<fracepsilon x^22$$ whenever $0<x<delta$. Similar inequality holds when $-delta <x<0$. It thus follows that the first limit in $(1)$ above is $0$. Thus the desired limit is $f(0)/2$.
answered yesterday
Paramanand Singh
45k553142
answered yesterday
Paramanand Singh
45k553142
answered yesterday
Paramanand Singh
45k553142
answered yesterday
Paramanand Singh
45k553142
45k553142
add a comment |Â
add a comment |Â
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