$limlimits_{ntoinfty}fleft(frac{x}{n}right)=0$ for every $x > 0$. Prove $limlimits_{x to 0}f(x)=0$
$begingroup$
Function $f: (0, infty) to mathbb{R}$ is continuous. For every positive $x$ we have $limlimits_{ntoinfty}fleft(frac{x}{n}right)=0$. Prove that $limlimits_{x to 0}f(x)=0$. I have tried to deduce something from definition of continuity, but with no effect.
limits continuity
edited Jan 7 at 22:40
rtybase
11.6k31534
asked Jan 7 at 22:20
user4201961user4201961
725411
$endgroup$
add a comment |
$begingroup$
Function $f: (0, infty) to mathbb{R}$ is continuous. For every positive $x$ we have $limlimits_{ntoinfty}fleft(frac{x}{n}right)=0$. Prove that $limlimits_{x to 0}f(x)=0$. I have tried to deduce something from definition of continuity, but with no effect.
limits continuity
edited Jan 7 at 22:40
rtybase
11.6k31534
asked Jan 7 at 22:20
user4201961user4201961
725411
$endgroup$
$begingroup$
What level course is this for? To me this is the kind of question that seems like a Baire Category Theorem thing.
$endgroup$
– Moya
Jan 7 at 22:41
$begingroup$
$n$ is an integer?
$endgroup$
– leonbloy
Jan 7 at 22:46
$begingroup$
But can you just not rewrite the limit such that $y=frac{x}{n}$ and so you get what you need by letting $y$ going to 0, which does happen as $n$ goes to infinity and $x$ is any positive real and f is continuous of course so the limit exists?
$endgroup$
– Marco Bellocchi
Jan 7 at 22:46
$begingroup$
@Moya introductory level
$endgroup$
– user4201961
Jan 7 at 22:48
$begingroup$
This I believe is just a bit more general than the following, you can proceed in a similar manner math.stackexchange.com/questions/3045776/…
$endgroup$
– Marco Bellocchi
Jan 7 at 22:59
add a comment |
$begingroup$
Function $f: (0, infty) to mathbb{R}$ is continuous. For every positive $x$ we have $limlimits_{ntoinfty}fleft(frac{x}{n}right)=0$. Prove that $limlimits_{x to 0}f(x)=0$. I have tried to deduce something from definition of continuity, but with no effect.
limits continuity
edited Jan 7 at 22:40
rtybase
11.6k31534
asked Jan 7 at 22:20
user4201961user4201961
725411
$endgroup$
Function $f: (0, infty) to mathbb{R}$ is continuous. For every positive $x$ we have $limlimits_{ntoinfty}fleft(frac{x}{n}right)=0$. Prove that $limlimits_{x to 0}f(x)=0$. I have tried to deduce something from definition of continuity, but with no effect.
limits continuity
limits continuity
edited Jan 7 at 22:40
rtybase
11.6k31534
asked Jan 7 at 22:20
user4201961user4201961
725411
edited Jan 7 at 22:40
rtybase
11.6k31534
asked Jan 7 at 22:20
user4201961user4201961
725411
edited Jan 7 at 22:40
rtybase
11.6k31534
edited Jan 7 at 22:40
rtybase
11.6k31534
edited Jan 7 at 22:40
rtybase
11.6k31534
11.6k31534
asked Jan 7 at 22:20
user4201961user4201961
725411
asked Jan 7 at 22:20
user4201961user4201961
725411
asked Jan 7 at 22:20
user4201961user4201961
725411
725411
$begingroup$
What level course is this for? To me this is the kind of question that seems like a Baire Category Theorem thing.
$endgroup$
– Moya
Jan 7 at 22:41
$begingroup$
$n$ is an integer?
$endgroup$
– leonbloy
Jan 7 at 22:46
$begingroup$
But can you just not rewrite the limit such that $y=frac{x}{n}$ and so you get what you need by letting $y$ going to 0, which does happen as $n$ goes to infinity and $x$ is any positive real and f is continuous of course so the limit exists?
$endgroup$
– Marco Bellocchi
Jan 7 at 22:46
$begingroup$
@Moya introductory level
$endgroup$
– user4201961
Jan 7 at 22:48
$begingroup$
This I believe is just a bit more general than the following, you can proceed in a similar manner math.stackexchange.com/questions/3045776/…
$endgroup$
– Marco Bellocchi
Jan 7 at 22:59
add a comment |
$begingroup$
What level course is this for? To me this is the kind of question that seems like a Baire Category Theorem thing.
$endgroup$
– Moya
Jan 7 at 22:41
$begingroup$
$n$ is an integer?
$endgroup$
– leonbloy
Jan 7 at 22:46
$begingroup$
But can you just not rewrite the limit such that $y=frac{x}{n}$ and so you get what you need by letting $y$ going to 0, which does happen as $n$ goes to infinity and $x$ is any positive real and f is continuous of course so the limit exists?
$endgroup$
– Marco Bellocchi
Jan 7 at 22:46
$begingroup$
@Moya introductory level
$endgroup$
– user4201961
Jan 7 at 22:48
$begingroup$
This I believe is just a bit more general than the following, you can proceed in a similar manner math.stackexchange.com/questions/3045776/…
$endgroup$
– Marco Bellocchi
Jan 7 at 22:59
$begingroup$
What level course is this for? To me this is the kind of question that seems like a Baire Category Theorem thing.
$endgroup$
– Moya
Jan 7 at 22:41
$begingroup$
What level course is this for? To me this is the kind of question that seems like a Baire Category Theorem thing.
$endgroup$
– Moya
Jan 7 at 22:41
$begingroup$
$n$ is an integer?
$endgroup$
– leonbloy
Jan 7 at 22:46
$begingroup$
$n$ is an integer?
$endgroup$
– leonbloy
Jan 7 at 22:46
$begingroup$
But can you just not rewrite the limit such that $y=frac{x}{n}$ and so you get what you need by letting $y$ going to 0, which does happen as $n$ goes to infinity and $x$ is any positive real and f is continuous of course so the limit exists?
$endgroup$
– Marco Bellocchi
Jan 7 at 22:46
$begingroup$
But can you just not rewrite the limit such that $y=frac{x}{n}$ and so you get what you need by letting $y$ going to 0, which does happen as $n$ goes to infinity and $x$ is any positive real and f is continuous of course so the limit exists?
$endgroup$
– Marco Bellocchi
Jan 7 at 22:46
$begingroup$
@Moya introductory level
$endgroup$
– user4201961
Jan 7 at 22:48
$begingroup$
@Moya introductory level
$endgroup$
– user4201961
Jan 7 at 22:48
$begingroup$
This I believe is just a bit more general than the following, you can proceed in a similar manner math.stackexchange.com/questions/3045776/…
$endgroup$
– Marco Bellocchi
Jan 7 at 22:59
$begingroup$
This I believe is just a bit more general than the following, you can proceed in a similar manner math.stackexchange.com/questions/3045776/…
$endgroup$
– Marco Bellocchi
Jan 7 at 22:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The proof relies on the following lemma:
Let $I subset (0,infty)$ be a closed bounded interval. Let $U subset (0,infty)$ be an open subset accumulating at $0$. Then there exists some integer $N geq 2$ and some closed interval $J subset U$ such that $N cdot J subset I$.
Sketch of proof: take some $x in U$ with $x$ lower than the length of $I$ and the minimum of $I$ . For some integer $N geq 2$, $Nx in overset{circ}{I}$, thus, there exists some compact interval $J subset U$ such that $N cdot J subset I$.
Now, assume $f$ does not go to $0$ at $0$. Then, for some $epsilon > 0$, $U={|f| > epsilon}$ is an open subset of $(0,infty)$ accumulating at $0$.
By iterating the lemma, we can construct sequences of compact intervals $(I_n)$ and integers $(N_n)$ such that $N_{n+1}I_{n+1} subset I_{n-1}$, $N_n geq 2$.
Now, let $A_n=N_1 ldots N_n$, then $A_n rightarrow infty$ and $K_n=A_n cdot I_n$ is a non-increasing sequence of nonempty compact intervals in $I_0$. Thus, there exists $x$ that is in every $K_n$.
So, $x/A_n in I_n subset U$, for all $n$, thus $|f(x/A_n)| > epsilon$ for all $n$, a contradiction.
answered Jan 7 at 22:50
MindlackMindlack
4,910211
$endgroup$
$begingroup$
(edited) This looks correct, but I think you can remove some of the intermediate claims without affecting the argument. In particular, you don't need any information about $sup I_n$. Also, when you say the sequence $K_n$ is non-decreasing, I think you mean that it is decreasing, or in more familiar language, descending. And in the final sentence, "all $n$" should be "infinitely many $n$".
$endgroup$
– Chris Culter
Jan 7 at 23:41
$begingroup$
Thanks for the remarks, I edited. However, in the last sentence, I definitely meant « for all $n$ » and not « for infinitely many $n$», however this does not affect the result.
$endgroup$
– Mindlack
Jan 8 at 0:02
$begingroup$
Reminds me of the Axiom of Archimedes.
$endgroup$
– marty cohen
Jan 8 at 0:49
add a comment |
$begingroup$
This is a standard application of Baire Category Theorem (BCT). Since $(0,infty)=cup_n {x:|f(frac x k)| leqepsilon ,forall k leq n}$ there exists $n$ such that ${x:|f(frac x k)| leq epsilon , forall k geq n}$ contains some open interval $(a,b)$. [ Because $(0,infty)$, being an open subset of $mathbb R$ has an equivalent complete metric so BCT applies]. This gives UNIFORM convergence of $f(frac x n)$ to $0$ on some open interval from which the result follows easily. [ Details: let $0<delta <b-a$ and $delta < frac a n$. Then, for any $x <delta $ the interval $(frac a x,frac b x)$ has length exceeding $1$, so it contains an interger $k$. Hence $kx in (a,b)$. Also, $k >frac a x > frac a {delta} > n$ so $|f(x)|=|f(frac {kx} k)| <epsilon$].
answered Jan 8 at 0:14
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
$endgroup$
$begingroup$
Shouldn’t it be $bigcup_n{x,,forall kgeq n,, |f(x/k)| leq epsilon}$?
$endgroup$
– Mindlack
Jan 8 at 0:54
$begingroup$
@Mindlack You are right. That was a silly mistake I made.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 5:01
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
The proof relies on the following lemma:
Let $I subset (0,infty)$ be a closed bounded interval. Let $U subset (0,infty)$ be an open subset accumulating at $0$. Then there exists some integer $N geq 2$ and some closed interval $J subset U$ such that $N cdot J subset I$.
Sketch of proof: take some $x in U$ with $x$ lower than the length of $I$ and the minimum of $I$ . For some integer $N geq 2$, $Nx in overset{circ}{I}$, thus, there exists some compact interval $J subset U$ such that $N cdot J subset I$.
Now, assume $f$ does not go to $0$ at $0$. Then, for some $epsilon > 0$, $U={|f| > epsilon}$ is an open subset of $(0,infty)$ accumulating at $0$.
By iterating the lemma, we can construct sequences of compact intervals $(I_n)$ and integers $(N_n)$ such that $N_{n+1}I_{n+1} subset I_{n-1}$, $N_n geq 2$.
Now, let $A_n=N_1 ldots N_n$, then $A_n rightarrow infty$ and $K_n=A_n cdot I_n$ is a non-increasing sequence of nonempty compact intervals in $I_0$. Thus, there exists $x$ that is in every $K_n$.
So, $x/A_n in I_n subset U$, for all $n$, thus $|f(x/A_n)| > epsilon$ for all $n$, a contradiction.
answered Jan 7 at 22:50
MindlackMindlack
4,910211
$endgroup$
$begingroup$
(edited) This looks correct, but I think you can remove some of the intermediate claims without affecting the argument. In particular, you don't need any information about $sup I_n$. Also, when you say the sequence $K_n$ is non-decreasing, I think you mean that it is decreasing, or in more familiar language, descending. And in the final sentence, "all $n$" should be "infinitely many $n$".
$endgroup$
– Chris Culter
Jan 7 at 23:41
$begingroup$
Thanks for the remarks, I edited. However, in the last sentence, I definitely meant « for all $n$ » and not « for infinitely many $n$», however this does not affect the result.
$endgroup$
– Mindlack
Jan 8 at 0:02
$begingroup$
Reminds me of the Axiom of Archimedes.
$endgroup$
– marty cohen
Jan 8 at 0:49
add a comment |
$begingroup$
The proof relies on the following lemma:
Let $I subset (0,infty)$ be a closed bounded interval. Let $U subset (0,infty)$ be an open subset accumulating at $0$. Then there exists some integer $N geq 2$ and some closed interval $J subset U$ such that $N cdot J subset I$.
Sketch of proof: take some $x in U$ with $x$ lower than the length of $I$ and the minimum of $I$ . For some integer $N geq 2$, $Nx in overset{circ}{I}$, thus, there exists some compact interval $J subset U$ such that $N cdot J subset I$.
Now, assume $f$ does not go to $0$ at $0$. Then, for some $epsilon > 0$, $U={|f| > epsilon}$ is an open subset of $(0,infty)$ accumulating at $0$.
By iterating the lemma, we can construct sequences of compact intervals $(I_n)$ and integers $(N_n)$ such that $N_{n+1}I_{n+1} subset I_{n-1}$, $N_n geq 2$.
Now, let $A_n=N_1 ldots N_n$, then $A_n rightarrow infty$ and $K_n=A_n cdot I_n$ is a non-increasing sequence of nonempty compact intervals in $I_0$. Thus, there exists $x$ that is in every $K_n$.
So, $x/A_n in I_n subset U$, for all $n$, thus $|f(x/A_n)| > epsilon$ for all $n$, a contradiction.
answered Jan 7 at 22:50
MindlackMindlack
4,910211
$endgroup$
$begingroup$
(edited) This looks correct, but I think you can remove some of the intermediate claims without affecting the argument. In particular, you don't need any information about $sup I_n$. Also, when you say the sequence $K_n$ is non-decreasing, I think you mean that it is decreasing, or in more familiar language, descending. And in the final sentence, "all $n$" should be "infinitely many $n$".
$endgroup$
– Chris Culter
Jan 7 at 23:41
$begingroup$
Thanks for the remarks, I edited. However, in the last sentence, I definitely meant « for all $n$ » and not « for infinitely many $n$», however this does not affect the result.
$endgroup$
– Mindlack
Jan 8 at 0:02
$begingroup$
Reminds me of the Axiom of Archimedes.
$endgroup$
– marty cohen
Jan 8 at 0:49
add a comment |
$begingroup$
The proof relies on the following lemma:
Let $I subset (0,infty)$ be a closed bounded interval. Let $U subset (0,infty)$ be an open subset accumulating at $0$. Then there exists some integer $N geq 2$ and some closed interval $J subset U$ such that $N cdot J subset I$.
Sketch of proof: take some $x in U$ with $x$ lower than the length of $I$ and the minimum of $I$ . For some integer $N geq 2$, $Nx in overset{circ}{I}$, thus, there exists some compact interval $J subset U$ such that $N cdot J subset I$.
Now, assume $f$ does not go to $0$ at $0$. Then, for some $epsilon > 0$, $U={|f| > epsilon}$ is an open subset of $(0,infty)$ accumulating at $0$.
By iterating the lemma, we can construct sequences of compact intervals $(I_n)$ and integers $(N_n)$ such that $N_{n+1}I_{n+1} subset I_{n-1}$, $N_n geq 2$.
Now, let $A_n=N_1 ldots N_n$, then $A_n rightarrow infty$ and $K_n=A_n cdot I_n$ is a non-increasing sequence of nonempty compact intervals in $I_0$. Thus, there exists $x$ that is in every $K_n$.
So, $x/A_n in I_n subset U$, for all $n$, thus $|f(x/A_n)| > epsilon$ for all $n$, a contradiction.
answered Jan 7 at 22:50
MindlackMindlack
4,910211
$endgroup$
The proof relies on the following lemma:
Let $I subset (0,infty)$ be a closed bounded interval. Let $U subset (0,infty)$ be an open subset accumulating at $0$. Then there exists some integer $N geq 2$ and some closed interval $J subset U$ such that $N cdot J subset I$.
Sketch of proof: take some $x in U$ with $x$ lower than the length of $I$ and the minimum of $I$ . For some integer $N geq 2$, $Nx in overset{circ}{I}$, thus, there exists some compact interval $J subset U$ such that $N cdot J subset I$.
Now, assume $f$ does not go to $0$ at $0$. Then, for some $epsilon > 0$, $U={|f| > epsilon}$ is an open subset of $(0,infty)$ accumulating at $0$.
By iterating the lemma, we can construct sequences of compact intervals $(I_n)$ and integers $(N_n)$ such that $N_{n+1}I_{n+1} subset I_{n-1}$, $N_n geq 2$.
Now, let $A_n=N_1 ldots N_n$, then $A_n rightarrow infty$ and $K_n=A_n cdot I_n$ is a non-increasing sequence of nonempty compact intervals in $I_0$. Thus, there exists $x$ that is in every $K_n$.
So, $x/A_n in I_n subset U$, for all $n$, thus $|f(x/A_n)| > epsilon$ for all $n$, a contradiction.
answered Jan 7 at 22:50
MindlackMindlack
4,910211
edited Jan 8 at 0:00
answered Jan 7 at 22:50
MindlackMindlack
4,910211
answered Jan 7 at 22:50
MindlackMindlack
4,910211
answered Jan 7 at 22:50
MindlackMindlack
4,910211
4,910211
$begingroup$
(edited) This looks correct, but I think you can remove some of the intermediate claims without affecting the argument. In particular, you don't need any information about $sup I_n$. Also, when you say the sequence $K_n$ is non-decreasing, I think you mean that it is decreasing, or in more familiar language, descending. And in the final sentence, "all $n$" should be "infinitely many $n$".
$endgroup$
– Chris Culter
Jan 7 at 23:41
$begingroup$
Thanks for the remarks, I edited. However, in the last sentence, I definitely meant « for all $n$ » and not « for infinitely many $n$», however this does not affect the result.
$endgroup$
– Mindlack
Jan 8 at 0:02
$begingroup$
Reminds me of the Axiom of Archimedes.
$endgroup$
– marty cohen
Jan 8 at 0:49
add a comment |
$begingroup$
(edited) This looks correct, but I think you can remove some of the intermediate claims without affecting the argument. In particular, you don't need any information about $sup I_n$. Also, when you say the sequence $K_n$ is non-decreasing, I think you mean that it is decreasing, or in more familiar language, descending. And in the final sentence, "all $n$" should be "infinitely many $n$".
$endgroup$
– Chris Culter
Jan 7 at 23:41
$begingroup$
Thanks for the remarks, I edited. However, in the last sentence, I definitely meant « for all $n$ » and not « for infinitely many $n$», however this does not affect the result.
$endgroup$
– Mindlack
Jan 8 at 0:02
$begingroup$
Reminds me of the Axiom of Archimedes.
$endgroup$
– marty cohen
Jan 8 at 0:49
$begingroup$
(edited) This looks correct, but I think you can remove some of the intermediate claims without affecting the argument. In particular, you don't need any information about $sup I_n$. Also, when you say the sequence $K_n$ is non-decreasing, I think you mean that it is decreasing, or in more familiar language, descending. And in the final sentence, "all $n$" should be "infinitely many $n$".
$endgroup$
– Chris Culter
Jan 7 at 23:41
$begingroup$
(edited) This looks correct, but I think you can remove some of the intermediate claims without affecting the argument. In particular, you don't need any information about $sup I_n$. Also, when you say the sequence $K_n$ is non-decreasing, I think you mean that it is decreasing, or in more familiar language, descending. And in the final sentence, "all $n$" should be "infinitely many $n$".
$endgroup$
– Chris Culter
Jan 7 at 23:41
$begingroup$
Thanks for the remarks, I edited. However, in the last sentence, I definitely meant « for all $n$ » and not « for infinitely many $n$», however this does not affect the result.
$endgroup$
– Mindlack
Jan 8 at 0:02
$begingroup$
Thanks for the remarks, I edited. However, in the last sentence, I definitely meant « for all $n$ » and not « for infinitely many $n$», however this does not affect the result.
$endgroup$
– Mindlack
Jan 8 at 0:02
$begingroup$
Reminds me of the Axiom of Archimedes.
$endgroup$
– marty cohen
Jan 8 at 0:49
$begingroup$
Reminds me of the Axiom of Archimedes.
$endgroup$
– marty cohen
Jan 8 at 0:49
add a comment |
$begingroup$
This is a standard application of Baire Category Theorem (BCT). Since $(0,infty)=cup_n {x:|f(frac x k)| leqepsilon ,forall k leq n}$ there exists $n$ such that ${x:|f(frac x k)| leq epsilon , forall k geq n}$ contains some open interval $(a,b)$. [ Because $(0,infty)$, being an open subset of $mathbb R$ has an equivalent complete metric so BCT applies]. This gives UNIFORM convergence of $f(frac x n)$ to $0$ on some open interval from which the result follows easily. [ Details: let $0<delta <b-a$ and $delta < frac a n$. Then, for any $x <delta $ the interval $(frac a x,frac b x)$ has length exceeding $1$, so it contains an interger $k$. Hence $kx in (a,b)$. Also, $k >frac a x > frac a {delta} > n$ so $|f(x)|=|f(frac {kx} k)| <epsilon$].
answered Jan 8 at 0:14
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
$endgroup$
$begingroup$
Shouldn’t it be $bigcup_n{x,,forall kgeq n,, |f(x/k)| leq epsilon}$?
$endgroup$
– Mindlack
Jan 8 at 0:54
$begingroup$
@Mindlack You are right. That was a silly mistake I made.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 5:01
add a comment |
$begingroup$
This is a standard application of Baire Category Theorem (BCT). Since $(0,infty)=cup_n {x:|f(frac x k)| leqepsilon ,forall k leq n}$ there exists $n$ such that ${x:|f(frac x k)| leq epsilon , forall k geq n}$ contains some open interval $(a,b)$. [ Because $(0,infty)$, being an open subset of $mathbb R$ has an equivalent complete metric so BCT applies]. This gives UNIFORM convergence of $f(frac x n)$ to $0$ on some open interval from which the result follows easily. [ Details: let $0<delta <b-a$ and $delta < frac a n$. Then, for any $x <delta $ the interval $(frac a x,frac b x)$ has length exceeding $1$, so it contains an interger $k$. Hence $kx in (a,b)$. Also, $k >frac a x > frac a {delta} > n$ so $|f(x)|=|f(frac {kx} k)| <epsilon$].
answered Jan 8 at 0:14
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
$endgroup$
$begingroup$
Shouldn’t it be $bigcup_n{x,,forall kgeq n,, |f(x/k)| leq epsilon}$?
$endgroup$
– Mindlack
Jan 8 at 0:54
$begingroup$
@Mindlack You are right. That was a silly mistake I made.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 5:01
add a comment |
$begingroup$
This is a standard application of Baire Category Theorem (BCT). Since $(0,infty)=cup_n {x:|f(frac x k)| leqepsilon ,forall k leq n}$ there exists $n$ such that ${x:|f(frac x k)| leq epsilon , forall k geq n}$ contains some open interval $(a,b)$. [ Because $(0,infty)$, being an open subset of $mathbb R$ has an equivalent complete metric so BCT applies]. This gives UNIFORM convergence of $f(frac x n)$ to $0$ on some open interval from which the result follows easily. [ Details: let $0<delta <b-a$ and $delta < frac a n$. Then, for any $x <delta $ the interval $(frac a x,frac b x)$ has length exceeding $1$, so it contains an interger $k$. Hence $kx in (a,b)$. Also, $k >frac a x > frac a {delta} > n$ so $|f(x)|=|f(frac {kx} k)| <epsilon$].
answered Jan 8 at 0:14
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
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This is a standard application of Baire Category Theorem (BCT). Since $(0,infty)=cup_n {x:|f(frac x k)| leqepsilon ,forall k leq n}$ there exists $n$ such that ${x:|f(frac x k)| leq epsilon , forall k geq n}$ contains some open interval $(a,b)$. [ Because $(0,infty)$, being an open subset of $mathbb R$ has an equivalent complete metric so BCT applies]. This gives UNIFORM convergence of $f(frac x n)$ to $0$ on some open interval from which the result follows easily. [ Details: let $0<delta <b-a$ and $delta < frac a n$. Then, for any $x <delta $ the interval $(frac a x,frac b x)$ has length exceeding $1$, so it contains an interger $k$. Hence $kx in (a,b)$. Also, $k >frac a x > frac a {delta} > n$ so $|f(x)|=|f(frac {kx} k)| <epsilon$].
answered Jan 8 at 0:14
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
edited Jan 8 at 5:13
answered Jan 8 at 0:14
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
answered Jan 8 at 0:14
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
answered Jan 8 at 0:14
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
74.4k53270
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Shouldn’t it be $bigcup_n{x,,forall kgeq n,, |f(x/k)| leq epsilon}$?
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– Mindlack
Jan 8 at 0:54
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@Mindlack You are right. That was a silly mistake I made.
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– Kavi Rama Murthy
Jan 8 at 5:01
add a comment |
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Shouldn’t it be $bigcup_n{x,,forall kgeq n,, |f(x/k)| leq epsilon}$?
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– Mindlack
Jan 8 at 0:54
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@Mindlack You are right. That was a silly mistake I made.
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– Kavi Rama Murthy
Jan 8 at 5:01
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Shouldn’t it be $bigcup_n{x,,forall kgeq n,, |f(x/k)| leq epsilon}$?
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– Mindlack
Jan 8 at 0:54
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Shouldn’t it be $bigcup_n{x,,forall kgeq n,, |f(x/k)| leq epsilon}$?
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– Mindlack
Jan 8 at 0:54
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@Mindlack You are right. That was a silly mistake I made.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 5:01
$begingroup$
@Mindlack You are right. That was a silly mistake I made.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 5:01
add a comment |
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What level course is this for? To me this is the kind of question that seems like a Baire Category Theorem thing.
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– Moya
Jan 7 at 22:41
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$n$ is an integer?
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– leonbloy
Jan 7 at 22:46
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But can you just not rewrite the limit such that $y=frac{x}{n}$ and so you get what you need by letting $y$ going to 0, which does happen as $n$ goes to infinity and $x$ is any positive real and f is continuous of course so the limit exists?
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– Marco Bellocchi
Jan 7 at 22:46
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@Moya introductory level
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– user4201961
Jan 7 at 22:48
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This I believe is just a bit more general than the following, you can proceed in a similar manner math.stackexchange.com/questions/3045776/…
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– Marco Bellocchi
Jan 7 at 22:59