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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 Aug 6 at 6:40
Asaf Karagila♦
292k31403733
asked Aug 6 at 0:02
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 Aug 6 at 6:40
Asaf Karagila♦
292k31403733
asked Aug 6 at 0:02
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 Aug 6 at 6:40
Asaf Karagila♦
292k31403733
asked Aug 6 at 0:02
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 Aug 6 at 6:40
Asaf Karagila♦
292k31403733
asked Aug 6 at 0:02
Konstantinos Zafeiris
1089
edited Aug 6 at 6:40
Asaf Karagila♦
292k31403733
edited Aug 6 at 6:40
Asaf Karagila♦
292k31403733
edited Aug 6 at 6:40
Asaf Karagila♦
292k31403733
292k31403733
asked Aug 6 at 0:02
Konstantinos Zafeiris
1089
asked Aug 6 at 0:02
Konstantinos Zafeiris
1089
asked Aug 6 at 0:02
Konstantinos Zafeiris
1089
1089
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
7
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 Aug 6 at 0:32
Ted Shifrin
59.6k44387
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
Aug 6 at 0:56
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
Aug 6 at 0:59
Ok, now I see, thank you.
– Konstantinos Zafeiris
Aug 6 at 1:00
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 Aug 6 at 0:32
Martin Argerami
116k1071164
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
Aug 6 at 0:43
@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.
– user580373
Aug 6 at 0:48
@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
Aug 6 at 0:50
@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.
– user580373
Aug 6 at 0:52
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– user580373
Aug 6 at 0:59
 |Â
show 2 more comments
up vote
5
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 Aug 6 at 1:09
Clement C.
47.2k33682
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
Aug 6 at 4:41
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
Aug 6 at 4:52
@TedShifrin Yes, good point.
– Clement C.
Aug 6 at 4:57
add a comment |Â
up vote
1
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 Aug 6 at 4:47
Paramanand Singh
45.3k553142
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
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 Aug 6 at 0:32
Ted Shifrin
59.6k44387
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
Aug 6 at 0:56
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
Aug 6 at 0:59
Ok, now I see, thank you.
– Konstantinos Zafeiris
Aug 6 at 1:00
add a comment |Â
up vote
7
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 Aug 6 at 0:32
Ted Shifrin
59.6k44387
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
Aug 6 at 0:56
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
Aug 6 at 0:59
Ok, now I see, thank you.
– Konstantinos Zafeiris
Aug 6 at 1:00
add a comment |Â
up vote
7
down vote
accepted
up vote
7
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 Aug 6 at 0:32
Ted Shifrin
59.6k44387
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 Aug 6 at 0:32
Ted Shifrin
59.6k44387
answered Aug 6 at 0:32
Ted Shifrin
59.6k44387
answered Aug 6 at 0:32
Ted Shifrin
59.6k44387
answered Aug 6 at 0:32
Ted Shifrin
59.6k44387
59.6k44387
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
Aug 6 at 0:56
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
Aug 6 at 0:59
Ok, now I see, thank you.
– Konstantinos Zafeiris
Aug 6 at 1:00
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
Aug 6 at 0:56
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
Aug 6 at 0:59
Ok, now I see, thank you.
– Konstantinos Zafeiris
Aug 6 at 1:00
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
Aug 6 at 0:56
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
Aug 6 at 0:56
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
Aug 6 at 0:59
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
Aug 6 at 0:59
Ok, now I see, thank you.
– Konstantinos Zafeiris
Aug 6 at 1:00
Ok, now I see, thank you.
– Konstantinos Zafeiris
Aug 6 at 1:00
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 Aug 6 at 0:32
Martin Argerami
116k1071164
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
Aug 6 at 0:43
@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.
– user580373
Aug 6 at 0:48
@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
Aug 6 at 0:50
@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.
– user580373
Aug 6 at 0:52
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– user580373
Aug 6 at 0:59
 |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 Aug 6 at 0:32
Martin Argerami
116k1071164
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
Aug 6 at 0:43
@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.
– user580373
Aug 6 at 0:48
@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
Aug 6 at 0:50
@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.
– user580373
Aug 6 at 0:52
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– user580373
Aug 6 at 0:59
 |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 Aug 6 at 0:32
Martin Argerami
116k1071164
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 Aug 6 at 0:32
Martin Argerami
116k1071164
answered Aug 6 at 0:32
Martin Argerami
116k1071164
answered Aug 6 at 0:32
Martin Argerami
116k1071164
answered Aug 6 at 0:32
Martin Argerami
116k1071164
116k1071164
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
Aug 6 at 0:43
@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.
– user580373
Aug 6 at 0:48
@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
Aug 6 at 0:50
@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.
– user580373
Aug 6 at 0:52
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– user580373
Aug 6 at 0:59
 |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
Aug 6 at 0:43
@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.
– user580373
Aug 6 at 0:48
@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
Aug 6 at 0:50
@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.
– user580373
Aug 6 at 0:52
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– user580373
Aug 6 at 0:59
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
Aug 6 at 0:43
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
Aug 6 at 0:43
@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.
– user580373
Aug 6 at 0:48
@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.
– user580373
Aug 6 at 0:48
@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
Aug 6 at 0:50
@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
Aug 6 at 0:50
@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.
– user580373
Aug 6 at 0:52
@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.
– user580373
Aug 6 at 0:52
1
1
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– user580373
Aug 6 at 0:59
@Sudix Inside the integral is $f(t)t$, not $f(t)$.
– user580373
Aug 6 at 0:59
 |Â
show 2 more comments
up vote
5
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 Aug 6 at 1:09
Clement C.
47.2k33682
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
Aug 6 at 4:41
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
Aug 6 at 4:52
@TedShifrin Yes, good point.
– Clement C.
Aug 6 at 4:57
add a comment |Â
up vote
5
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 Aug 6 at 1:09
Clement C.
47.2k33682
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
Aug 6 at 4:41
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
Aug 6 at 4:52
@TedShifrin Yes, good point.
– Clement C.
Aug 6 at 4:57
add a comment |Â
up vote
5
down vote
up vote
5
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 Aug 6 at 1:09
Clement C.
47.2k33682
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 Aug 6 at 1:09
Clement C.
47.2k33682
edited Aug 6 at 1:37
answered Aug 6 at 1:09
Clement C.
47.2k33682
answered Aug 6 at 1:09
Clement C.
47.2k33682
answered Aug 6 at 1:09
Clement C.
47.2k33682
47.2k33682
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
Aug 6 at 4:41
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
Aug 6 at 4:52
@TedShifrin Yes, good point.
– Clement C.
Aug 6 at 4:57
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
Aug 6 at 4:41
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
Aug 6 at 4:52
@TedShifrin Yes, good point.
– Clement C.
Aug 6 at 4:57
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
Aug 6 at 4:41
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
Aug 6 at 4:41
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
Aug 6 at 4:52
@TedShifrin: I used the same approach as you mention in your comment and I saw your comment later.
– Paramanand Singh
Aug 6 at 4:52
@TedShifrin Yes, good point.
– Clement C.
Aug 6 at 4:57
@TedShifrin Yes, good point.
– Clement C.
Aug 6 at 4:57
add a comment |Â
up vote
1
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 Aug 6 at 4:47
Paramanand Singh
45.3k553142
add a comment |Â
up vote
1
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 Aug 6 at 4:47
Paramanand Singh
45.3k553142
add a comment |Â
up vote
1
down vote
up vote
1
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 Aug 6 at 4:47
Paramanand Singh
45.3k553142
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 Aug 6 at 4:47
Paramanand Singh
45.3k553142
answered Aug 6 at 4:47
Paramanand Singh
45.3k553142
answered Aug 6 at 4:47
Paramanand Singh
45.3k553142
answered Aug 6 at 4:47
Paramanand Singh
45.3k553142
45.3k553142
add a comment |Â
add a comment |Â
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