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How can I prove this sequence of functions pointwisely converges to zero using formal argument?
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Let $f_n(x)$ be a sequence of functions defined on $ [0,1]$ by
$$ f_n(x)= begincases n text if $0< x < frac1n$, \ 0 text if $x=0$ or $ frac1nle x le 1$ endcases$$
How to prove the pointwise limit of $f_n(x)$ is $0$ using $epsilon-delta $ argument ?
real-analysis sequences-and-series epsilon-delta sequence-of-function
edited Jul 15 at 14:33
Martin Sleziak
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asked Jul 15 at 10:22
user577285
112
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up vote
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Let $f_n(x)$ be a sequence of functions defined on $ [0,1]$ by
$$ f_n(x)= begincases n text if $0< x < frac1n$, \ 0 text if $x=0$ or $ frac1nle x le 1$ endcases$$
How to prove the pointwise limit of $f_n(x)$ is $0$ using $epsilon-delta $ argument ?
real-analysis sequences-and-series epsilon-delta sequence-of-function
edited Jul 15 at 14:33
Martin Sleziak
43.5k6113259
asked Jul 15 at 10:22
user577285
112
You need an $epsilon$ - $N$ argument, i.e. you needs to show that for any given $epsilon$ and $x$ there is a $N$ such that $|f_n(x) - 0| < epsilon$ for all $n > N$. For this try to show that for any $x > 0$ you choose taking $n$ large enough then $f_n(x) = 0$.
– Winther
Jul 15 at 10:39
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up vote
2
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up vote
2
down vote
favorite
Let $f_n(x)$ be a sequence of functions defined on $ [0,1]$ by
$$ f_n(x)= begincases n text if $0< x < frac1n$, \ 0 text if $x=0$ or $ frac1nle x le 1$ endcases$$
How to prove the pointwise limit of $f_n(x)$ is $0$ using $epsilon-delta $ argument ?
real-analysis sequences-and-series epsilon-delta sequence-of-function
edited Jul 15 at 14:33
Martin Sleziak
43.5k6113259
asked Jul 15 at 10:22
user577285
112
Let $f_n(x)$ be a sequence of functions defined on $ [0,1]$ by
$$ f_n(x)= begincases n text if $0< x < frac1n$, \ 0 text if $x=0$ or $ frac1nle x le 1$ endcases$$
How to prove the pointwise limit of $f_n(x)$ is $0$ using $epsilon-delta $ argument ?
real-analysis sequences-and-series epsilon-delta sequence-of-function
edited Jul 15 at 14:33
Martin Sleziak
43.5k6113259
asked Jul 15 at 10:22
user577285
112
edited Jul 15 at 14:33
Martin Sleziak
43.5k6113259
edited Jul 15 at 14:33
Martin Sleziak
43.5k6113259
edited Jul 15 at 14:33
Martin Sleziak
43.5k6113259
43.5k6113259
asked Jul 15 at 10:22
user577285
112
asked Jul 15 at 10:22
user577285
112
asked Jul 15 at 10:22
user577285
112
112
You need an $epsilon$ - $N$ argument, i.e. you needs to show that for any given $epsilon$ and $x$ there is a $N$ such that $|f_n(x) - 0| < epsilon$ for all $n > N$. For this try to show that for any $x > 0$ you choose taking $n$ large enough then $f_n(x) = 0$.
– Winther
Jul 15 at 10:39
add a comment |Â
You need an $epsilon$ - $N$ argument, i.e. you needs to show that for any given $epsilon$ and $x$ there is a $N$ such that $|f_n(x) - 0| < epsilon$ for all $n > N$. For this try to show that for any $x > 0$ you choose taking $n$ large enough then $f_n(x) = 0$.
– Winther
Jul 15 at 10:39
You need an $epsilon$ - $N$ argument, i.e. you needs to show that for any given $epsilon$ and $x$ there is a $N$ such that $|f_n(x) - 0| < epsilon$ for all $n > N$. For this try to show that for any $x > 0$ you choose taking $n$ large enough then $f_n(x) = 0$.
– Winther
Jul 15 at 10:39
You need an $epsilon$ - $N$ argument, i.e. you needs to show that for any given $epsilon$ and $x$ there is a $N$ such that $|f_n(x) - 0| < epsilon$ for all $n > N$. For this try to show that for any $x > 0$ you choose taking $n$ large enough then $f_n(x) = 0$.
– Winther
Jul 15 at 10:39
add a comment |Â
2 Answers
2
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up vote
3
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Hint:
Fix $x in (0,1]$ and pick $n_0 in mathbbN$ such that $frac1n_0 le x$. Then $f_n(x) = 0$ for all $n ge n_0$.
answered Jul 15 at 10:47
mechanodroid
22.3k52041
add a comment |Â
up vote
1
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$textbfSolution$:
Fix $x in (0,1)$, as if $x=1$, then $f_n(1)=0$ for every $n$ and we have the claim and similarly for $x=0$. So now assume WLOG $x in (0,1)$, and let $epsilon > 0$. Then by archmedian's principle, there exists an $N in mathbbN$ such that beginalign frac1N < x endalign In particular, $x in (1/N,1) subset [1/N,1]$. Then beginalign f_N(x) = 0 endalign
Then we notice that for any $n geq N$, $frac1N geq frac1n$, so we have $(1/N,1) subset (1/n,1)$. In particular, this means for any $n geq N, f_n(x) = 0$. Then the rest follows from $0 < epsilon$.
answered Jul 15 at 16:57
Raymond Chu
1,03719
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Hint:
Fix $x in (0,1]$ and pick $n_0 in mathbbN$ such that $frac1n_0 le x$. Then $f_n(x) = 0$ for all $n ge n_0$.
answered Jul 15 at 10:47
mechanodroid
22.3k52041
add a comment |Â
up vote
3
down vote
Hint:
Fix $x in (0,1]$ and pick $n_0 in mathbbN$ such that $frac1n_0 le x$. Then $f_n(x) = 0$ for all $n ge n_0$.
answered Jul 15 at 10:47
mechanodroid
22.3k52041
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Hint:
Fix $x in (0,1]$ and pick $n_0 in mathbbN$ such that $frac1n_0 le x$. Then $f_n(x) = 0$ for all $n ge n_0$.
answered Jul 15 at 10:47
mechanodroid
22.3k52041
Hint:
Fix $x in (0,1]$ and pick $n_0 in mathbbN$ such that $frac1n_0 le x$. Then $f_n(x) = 0$ for all $n ge n_0$.
answered Jul 15 at 10:47
mechanodroid
22.3k52041
answered Jul 15 at 10:47
mechanodroid
22.3k52041
answered Jul 15 at 10:47
mechanodroid
22.3k52041
answered Jul 15 at 10:47
mechanodroid
22.3k52041
22.3k52041
add a comment |Â
add a comment |Â
up vote
1
down vote
$textbfSolution$:
Fix $x in (0,1)$, as if $x=1$, then $f_n(1)=0$ for every $n$ and we have the claim and similarly for $x=0$. So now assume WLOG $x in (0,1)$, and let $epsilon > 0$. Then by archmedian's principle, there exists an $N in mathbbN$ such that beginalign frac1N < x endalign In particular, $x in (1/N,1) subset [1/N,1]$. Then beginalign f_N(x) = 0 endalign
Then we notice that for any $n geq N$, $frac1N geq frac1n$, so we have $(1/N,1) subset (1/n,1)$. In particular, this means for any $n geq N, f_n(x) = 0$. Then the rest follows from $0 < epsilon$.
answered Jul 15 at 16:57
Raymond Chu
1,03719
add a comment |Â
up vote
1
down vote
$textbfSolution$:
Fix $x in (0,1)$, as if $x=1$, then $f_n(1)=0$ for every $n$ and we have the claim and similarly for $x=0$. So now assume WLOG $x in (0,1)$, and let $epsilon > 0$. Then by archmedian's principle, there exists an $N in mathbbN$ such that beginalign frac1N < x endalign In particular, $x in (1/N,1) subset [1/N,1]$. Then beginalign f_N(x) = 0 endalign
Then we notice that for any $n geq N$, $frac1N geq frac1n$, so we have $(1/N,1) subset (1/n,1)$. In particular, this means for any $n geq N, f_n(x) = 0$. Then the rest follows from $0 < epsilon$.
answered Jul 15 at 16:57
Raymond Chu
1,03719
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$textbfSolution$:
Fix $x in (0,1)$, as if $x=1$, then $f_n(1)=0$ for every $n$ and we have the claim and similarly for $x=0$. So now assume WLOG $x in (0,1)$, and let $epsilon > 0$. Then by archmedian's principle, there exists an $N in mathbbN$ such that beginalign frac1N < x endalign In particular, $x in (1/N,1) subset [1/N,1]$. Then beginalign f_N(x) = 0 endalign
Then we notice that for any $n geq N$, $frac1N geq frac1n$, so we have $(1/N,1) subset (1/n,1)$. In particular, this means for any $n geq N, f_n(x) = 0$. Then the rest follows from $0 < epsilon$.
answered Jul 15 at 16:57
Raymond Chu
1,03719
$textbfSolution$:
Fix $x in (0,1)$, as if $x=1$, then $f_n(1)=0$ for every $n$ and we have the claim and similarly for $x=0$. So now assume WLOG $x in (0,1)$, and let $epsilon > 0$. Then by archmedian's principle, there exists an $N in mathbbN$ such that beginalign frac1N < x endalign In particular, $x in (1/N,1) subset [1/N,1]$. Then beginalign f_N(x) = 0 endalign
Then we notice that for any $n geq N$, $frac1N geq frac1n$, so we have $(1/N,1) subset (1/n,1)$. In particular, this means for any $n geq N, f_n(x) = 0$. Then the rest follows from $0 < epsilon$.
answered Jul 15 at 16:57
Raymond Chu
1,03719
answered Jul 15 at 16:57
Raymond Chu
1,03719
answered Jul 15 at 16:57
Raymond Chu
1,03719
answered Jul 15 at 16:57
Raymond Chu
1,03719
1,03719
add a comment |Â
add a comment |Â
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You need an $epsilon$ - $N$ argument, i.e. you needs to show that for any given $epsilon$ and $x$ there is a $N$ such that $|f_n(x) - 0| < epsilon$ for all $n > N$. For this try to show that for any $x > 0$ you choose taking $n$ large enough then $f_n(x) = 0$.
– Winther
Jul 15 at 10:39