Let $f$ be a continuous function on $[0,infty)$ such that $lim_{xto infty}(f(x)+int_0^x f(t)dt)$ exists. Find...
Let $f$ be a continuous function on $[0,infty)$ such that $lim_{xto infty}(f(x)+int_0^x f(t)dt)$ exists. Find $lim_{xto infty}f(x)$.
Useful hints will work. Please give some nice hint to solve the problem.
integration limits functions continuity
edited Nov 29 at 21:00
Shaun
8,507113580
asked Feb 2 '17 at 14:32
Piyush Raut
577
|
show 3 more comments
Let $f$ be a continuous function on $[0,infty)$ such that $lim_{xto infty}(f(x)+int_0^x f(t)dt)$ exists. Find $lim_{xto infty}f(x)$.
Useful hints will work. Please give some nice hint to solve the problem.
integration limits functions continuity
edited Nov 29 at 21:00
Shaun
8,507113580
asked Feb 2 '17 at 14:32
Piyush Raut
577
Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
– xavierm02
Feb 2 '17 at 14:36
@xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
– Piyush Raut
Feb 2 '17 at 14:39
But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
– xavierm02
Feb 2 '17 at 14:42
I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
– Piyush Raut
Feb 2 '17 at 14:45
No I meant with $g$, under the assumption that the limit exists with $f$.
– xavierm02
Feb 2 '17 at 14:45
|
show 3 more comments
Let $f$ be a continuous function on $[0,infty)$ such that $lim_{xto infty}(f(x)+int_0^x f(t)dt)$ exists. Find $lim_{xto infty}f(x)$.
Useful hints will work. Please give some nice hint to solve the problem.
integration limits functions continuity
edited Nov 29 at 21:00
Shaun
8,507113580
asked Feb 2 '17 at 14:32
Piyush Raut
577
Let $f$ be a continuous function on $[0,infty)$ such that $lim_{xto infty}(f(x)+int_0^x f(t)dt)$ exists. Find $lim_{xto infty}f(x)$.
Useful hints will work. Please give some nice hint to solve the problem.
integration limits functions continuity
integration limits functions continuity
edited Nov 29 at 21:00
Shaun
8,507113580
asked Feb 2 '17 at 14:32
Piyush Raut
577
edited Nov 29 at 21:00
Shaun
8,507113580
asked Feb 2 '17 at 14:32
Piyush Raut
577
edited Nov 29 at 21:00
Shaun
8,507113580
edited Nov 29 at 21:00
Shaun
8,507113580
edited Nov 29 at 21:00
Shaun
8,507113580
8,507113580
asked Feb 2 '17 at 14:32
Piyush Raut
577
asked Feb 2 '17 at 14:32
Piyush Raut
577
asked Feb 2 '17 at 14:32
Piyush Raut
577
577
Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
– xavierm02
Feb 2 '17 at 14:36
@xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
– Piyush Raut
Feb 2 '17 at 14:39
But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
– xavierm02
Feb 2 '17 at 14:42
I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
– Piyush Raut
Feb 2 '17 at 14:45
No I meant with $g$, under the assumption that the limit exists with $f$.
– xavierm02
Feb 2 '17 at 14:45
|
show 3 more comments
Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
– xavierm02
Feb 2 '17 at 14:36
@xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
– Piyush Raut
Feb 2 '17 at 14:39
But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
– xavierm02
Feb 2 '17 at 14:42
I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
– Piyush Raut
Feb 2 '17 at 14:45
No I meant with $g$, under the assumption that the limit exists with $f$.
– xavierm02
Feb 2 '17 at 14:45
Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
– xavierm02
Feb 2 '17 at 14:36
Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
– xavierm02
Feb 2 '17 at 14:36
@xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
– Piyush Raut
Feb 2 '17 at 14:39
@xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
– Piyush Raut
Feb 2 '17 at 14:39
But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
– xavierm02
Feb 2 '17 at 14:42
But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
– xavierm02
Feb 2 '17 at 14:42
I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
– Piyush Raut
Feb 2 '17 at 14:45
I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
– Piyush Raut
Feb 2 '17 at 14:45
No I meant with $g$, under the assumption that the limit exists with $f$.
– xavierm02
Feb 2 '17 at 14:45
No I meant with $g$, under the assumption that the limit exists with $f$.
– xavierm02
Feb 2 '17 at 14:45
|
show 3 more comments
2 Answers
2
active
oldest
votes
Let $F(x) = int_0^x f(t)dt$. Then, we have $lim_{xto infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $epsilon>0$, there is $N>0$ such that for all $x>N$,
$$
(c-epsilon)e^x leq[e^x F(x)]' leq (c+epsilon)e^x.
$$
Now, integrating from $N$ to $t$ gives:
$$(c-epsilon)(e^t-e^N)leq e^tF(t) -e^NF(N) leq (c+epsilon)(e^t-e^N),quadforall t>N
$$and hence
$$
(c-epsilon)(1-e^{N-t})leq F(t) -e^{N-t}F(N) leq (c+epsilon)(1-e^{N-t}), quad forall t>N.
$$Taking $ttoinfty$ yields
$$c-epsilon leq liminf_{tto infty}F(t)leq limsup_{tto infty}F(t)leq c+epsilon.
$$Since $epsilon>0$ was arbitrary, we have $lim_{ttoinfty} F(t) =c$ and thus $lim_{ttoinfty} f(t) = 0$.
answered Nov 27 at 7:45
Song
4,225316
add a comment |
One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =int_{0}^{x}f(t),dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) to L$ as $xtoinfty$.
We have
begin{align}
lim_{xtoinfty} F(x) &=lim_{xtoinfty} frac{e^xF(x)} {e^x}notag\
&=lim_{xto infty} dfrac{e^x{F(x) +F'(x) }} {e^x}text{ (via L'Hospital's Rule)} notag\
&=lim_{xtoinfty} F(x) +F'(x)notag\
&=Lnotag
end{align}
and hence $$lim_{xto infty} f(x) =lim_{xtoinfty} F'(x) =lim_{xtoinfty} {F(x) + F'(x) }-F(x) =L-L=0$$
answered Nov 30 at 2:37
Paramanand Singh
48.9k555157
1
Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
– zhw.
Nov 30 at 17:51
@zhw : I should have added more details. Thanks for putting these details in your comment.
– Paramanand Singh
Nov 30 at 17:55
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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oldest
votes
Let $F(x) = int_0^x f(t)dt$. Then, we have $lim_{xto infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $epsilon>0$, there is $N>0$ such that for all $x>N$,
$$
(c-epsilon)e^x leq[e^x F(x)]' leq (c+epsilon)e^x.
$$
Now, integrating from $N$ to $t$ gives:
$$(c-epsilon)(e^t-e^N)leq e^tF(t) -e^NF(N) leq (c+epsilon)(e^t-e^N),quadforall t>N
$$and hence
$$
(c-epsilon)(1-e^{N-t})leq F(t) -e^{N-t}F(N) leq (c+epsilon)(1-e^{N-t}), quad forall t>N.
$$Taking $ttoinfty$ yields
$$c-epsilon leq liminf_{tto infty}F(t)leq limsup_{tto infty}F(t)leq c+epsilon.
$$Since $epsilon>0$ was arbitrary, we have $lim_{ttoinfty} F(t) =c$ and thus $lim_{ttoinfty} f(t) = 0$.
answered Nov 27 at 7:45
Song
4,225316
add a comment |
Let $F(x) = int_0^x f(t)dt$. Then, we have $lim_{xto infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $epsilon>0$, there is $N>0$ such that for all $x>N$,
$$
(c-epsilon)e^x leq[e^x F(x)]' leq (c+epsilon)e^x.
$$
Now, integrating from $N$ to $t$ gives:
$$(c-epsilon)(e^t-e^N)leq e^tF(t) -e^NF(N) leq (c+epsilon)(e^t-e^N),quadforall t>N
$$and hence
$$
(c-epsilon)(1-e^{N-t})leq F(t) -e^{N-t}F(N) leq (c+epsilon)(1-e^{N-t}), quad forall t>N.
$$Taking $ttoinfty$ yields
$$c-epsilon leq liminf_{tto infty}F(t)leq limsup_{tto infty}F(t)leq c+epsilon.
$$Since $epsilon>0$ was arbitrary, we have $lim_{ttoinfty} F(t) =c$ and thus $lim_{ttoinfty} f(t) = 0$.
answered Nov 27 at 7:45
Song
4,225316
add a comment |
Let $F(x) = int_0^x f(t)dt$. Then, we have $lim_{xto infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $epsilon>0$, there is $N>0$ such that for all $x>N$,
$$
(c-epsilon)e^x leq[e^x F(x)]' leq (c+epsilon)e^x.
$$
Now, integrating from $N$ to $t$ gives:
$$(c-epsilon)(e^t-e^N)leq e^tF(t) -e^NF(N) leq (c+epsilon)(e^t-e^N),quadforall t>N
$$and hence
$$
(c-epsilon)(1-e^{N-t})leq F(t) -e^{N-t}F(N) leq (c+epsilon)(1-e^{N-t}), quad forall t>N.
$$Taking $ttoinfty$ yields
$$c-epsilon leq liminf_{tto infty}F(t)leq limsup_{tto infty}F(t)leq c+epsilon.
$$Since $epsilon>0$ was arbitrary, we have $lim_{ttoinfty} F(t) =c$ and thus $lim_{ttoinfty} f(t) = 0$.
answered Nov 27 at 7:45
Song
4,225316
Let $F(x) = int_0^x f(t)dt$. Then, we have $lim_{xto infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $epsilon>0$, there is $N>0$ such that for all $x>N$,
$$
(c-epsilon)e^x leq[e^x F(x)]' leq (c+epsilon)e^x.
$$
Now, integrating from $N$ to $t$ gives:
$$(c-epsilon)(e^t-e^N)leq e^tF(t) -e^NF(N) leq (c+epsilon)(e^t-e^N),quadforall t>N
$$and hence
$$
(c-epsilon)(1-e^{N-t})leq F(t) -e^{N-t}F(N) leq (c+epsilon)(1-e^{N-t}), quad forall t>N.
$$Taking $ttoinfty$ yields
$$c-epsilon leq liminf_{tto infty}F(t)leq limsup_{tto infty}F(t)leq c+epsilon.
$$Since $epsilon>0$ was arbitrary, we have $lim_{ttoinfty} F(t) =c$ and thus $lim_{ttoinfty} f(t) = 0$.
answered Nov 27 at 7:45
Song
4,225316
answered Nov 27 at 7:45
Song
4,225316
answered Nov 27 at 7:45
Song
4,225316
answered Nov 27 at 7:45
Song
4,225316
4,225316
add a comment |
add a comment |
One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =int_{0}^{x}f(t),dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) to L$ as $xtoinfty$.
We have
begin{align}
lim_{xtoinfty} F(x) &=lim_{xtoinfty} frac{e^xF(x)} {e^x}notag\
&=lim_{xto infty} dfrac{e^x{F(x) +F'(x) }} {e^x}text{ (via L'Hospital's Rule)} notag\
&=lim_{xtoinfty} F(x) +F'(x)notag\
&=Lnotag
end{align}
and hence $$lim_{xto infty} f(x) =lim_{xtoinfty} F'(x) =lim_{xtoinfty} {F(x) + F'(x) }-F(x) =L-L=0$$
answered Nov 30 at 2:37
Paramanand Singh
48.9k555157
1
Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
– zhw.
Nov 30 at 17:51
@zhw : I should have added more details. Thanks for putting these details in your comment.
– Paramanand Singh
Nov 30 at 17:55
add a comment |
One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =int_{0}^{x}f(t),dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) to L$ as $xtoinfty$.
We have
begin{align}
lim_{xtoinfty} F(x) &=lim_{xtoinfty} frac{e^xF(x)} {e^x}notag\
&=lim_{xto infty} dfrac{e^x{F(x) +F'(x) }} {e^x}text{ (via L'Hospital's Rule)} notag\
&=lim_{xtoinfty} F(x) +F'(x)notag\
&=Lnotag
end{align}
and hence $$lim_{xto infty} f(x) =lim_{xtoinfty} F'(x) =lim_{xtoinfty} {F(x) + F'(x) }-F(x) =L-L=0$$
answered Nov 30 at 2:37
Paramanand Singh
48.9k555157
1
Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
– zhw.
Nov 30 at 17:51
@zhw : I should have added more details. Thanks for putting these details in your comment.
– Paramanand Singh
Nov 30 at 17:55
add a comment |
One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =int_{0}^{x}f(t),dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) to L$ as $xtoinfty$.
We have
begin{align}
lim_{xtoinfty} F(x) &=lim_{xtoinfty} frac{e^xF(x)} {e^x}notag\
&=lim_{xto infty} dfrac{e^x{F(x) +F'(x) }} {e^x}text{ (via L'Hospital's Rule)} notag\
&=lim_{xtoinfty} F(x) +F'(x)notag\
&=Lnotag
end{align}
and hence $$lim_{xto infty} f(x) =lim_{xtoinfty} F'(x) =lim_{xtoinfty} {F(x) + F'(x) }-F(x) =L-L=0$$
answered Nov 30 at 2:37
Paramanand Singh
48.9k555157
One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =int_{0}^{x}f(t),dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) to L$ as $xtoinfty$.
We have
begin{align}
lim_{xtoinfty} F(x) &=lim_{xtoinfty} frac{e^xF(x)} {e^x}notag\
&=lim_{xto infty} dfrac{e^x{F(x) +F'(x) }} {e^x}text{ (via L'Hospital's Rule)} notag\
&=lim_{xtoinfty} F(x) +F'(x)notag\
&=Lnotag
end{align}
and hence $$lim_{xto infty} f(x) =lim_{xtoinfty} F'(x) =lim_{xtoinfty} {F(x) + F'(x) }-F(x) =L-L=0$$
answered Nov 30 at 2:37
Paramanand Singh
48.9k555157
answered Nov 30 at 2:37
Paramanand Singh
48.9k555157
answered Nov 30 at 2:37
Paramanand Singh
48.9k555157
answered Nov 30 at 2:37
Paramanand Singh
48.9k555157
48.9k555157
1
Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
– zhw.
Nov 30 at 17:51
@zhw : I should have added more details. Thanks for putting these details in your comment.
– Paramanand Singh
Nov 30 at 17:55
add a comment |
1
Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
– zhw.
Nov 30 at 17:51
@zhw : I should have added more details. Thanks for putting these details in your comment.
– Paramanand Singh
Nov 30 at 17:55
1
1
Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
– zhw.
Nov 30 at 17:51
Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
– zhw.
Nov 30 at 17:51
@zhw : I should have added more details. Thanks for putting these details in your comment.
– Paramanand Singh
Nov 30 at 17:55
@zhw : I should have added more details. Thanks for putting these details in your comment.
– Paramanand Singh
Nov 30 at 17:55
add a comment |
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Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
– xavierm02
Feb 2 '17 at 14:36
@xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
– Piyush Raut
Feb 2 '17 at 14:39
But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
– xavierm02
Feb 2 '17 at 14:42
I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
– Piyush Raut
Feb 2 '17 at 14:45
No I meant with $g$, under the assumption that the limit exists with $f$.
– xavierm02
Feb 2 '17 at 14:45