Pointwise limit function of a piecewise function
$begingroup$
Find the pointwise limit function of:
$$f_n(x)=begin{cases}
0 & |x|> 1/n \
nx+1 & x in [-1/n, 0) \
1-nx & x in [0, 1/n]
end{cases} $$
I think that in the limit, if we fix a certain $x$, we get that:
$$lim _{n to infty} f_n(x)=begin{cases}
0 & |x|> 0 \
infty & x in [0, 0) \
-infty & x in [0, 0]
end{cases} $$
Where the second line is an empty statement, we rewrite this to:
$$lim _{n to infty} f_n(x)=begin{cases}
0 & |x|> 0 \
-infty & x =0
end{cases} $$
Did I do this correctly? I'm not sure what the formal argument should be to get rid of the half-open interval $[0,0)$
real-analysis limits functions proof-verification pointwise-convergence
asked Jan 7 at 12:28
Wesley StrikWesley Strik
1,635423
$endgroup$
add a comment |
$begingroup$
Find the pointwise limit function of:
$$f_n(x)=begin{cases}
0 & |x|> 1/n \
nx+1 & x in [-1/n, 0) \
1-nx & x in [0, 1/n]
end{cases} $$
I think that in the limit, if we fix a certain $x$, we get that:
$$lim _{n to infty} f_n(x)=begin{cases}
0 & |x|> 0 \
infty & x in [0, 0) \
-infty & x in [0, 0]
end{cases} $$
Where the second line is an empty statement, we rewrite this to:
$$lim _{n to infty} f_n(x)=begin{cases}
0 & |x|> 0 \
-infty & x =0
end{cases} $$
Did I do this correctly? I'm not sure what the formal argument should be to get rid of the half-open interval $[0,0)$
real-analysis limits functions proof-verification pointwise-convergence
asked Jan 7 at 12:28
Wesley StrikWesley Strik
1,635423
$endgroup$
add a comment |
$begingroup$
Find the pointwise limit function of:
$$f_n(x)=begin{cases}
0 & |x|> 1/n \
nx+1 & x in [-1/n, 0) \
1-nx & x in [0, 1/n]
end{cases} $$
I think that in the limit, if we fix a certain $x$, we get that:
$$lim _{n to infty} f_n(x)=begin{cases}
0 & |x|> 0 \
infty & x in [0, 0) \
-infty & x in [0, 0]
end{cases} $$
Where the second line is an empty statement, we rewrite this to:
$$lim _{n to infty} f_n(x)=begin{cases}
0 & |x|> 0 \
-infty & x =0
end{cases} $$
Did I do this correctly? I'm not sure what the formal argument should be to get rid of the half-open interval $[0,0)$
real-analysis limits functions proof-verification pointwise-convergence
asked Jan 7 at 12:28
Wesley StrikWesley Strik
1,635423
$endgroup$
Find the pointwise limit function of:
$$f_n(x)=begin{cases}
0 & |x|> 1/n \
nx+1 & x in [-1/n, 0) \
1-nx & x in [0, 1/n]
end{cases} $$
I think that in the limit, if we fix a certain $x$, we get that:
$$lim _{n to infty} f_n(x)=begin{cases}
0 & |x|> 0 \
infty & x in [0, 0) \
-infty & x in [0, 0]
end{cases} $$
Where the second line is an empty statement, we rewrite this to:
$$lim _{n to infty} f_n(x)=begin{cases}
0 & |x|> 0 \
-infty & x =0
end{cases} $$
Did I do this correctly? I'm not sure what the formal argument should be to get rid of the half-open interval $[0,0)$
real-analysis limits functions proof-verification pointwise-convergence
real-analysis limits functions proof-verification pointwise-convergence
asked Jan 7 at 12:28
Wesley StrikWesley Strik
1,635423
asked Jan 7 at 12:28
Wesley StrikWesley Strik
1,635423
asked Jan 7 at 12:28
Wesley StrikWesley Strik
1,635423
asked Jan 7 at 12:28
Wesley StrikWesley Strik
1,635423
asked Jan 7 at 12:28
Wesley StrikWesley Strik
1,635423
1,635423
add a comment |
add a comment |
3 Answers
3
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Your answer is not correct. If $x=0$ then $f_n(x)=1$ for all $n$ and the limit is $1$. If $x neq 0$ then $f_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.
answered Jan 7 at 12:32
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
$begingroup$
yes that makes sense
$endgroup$
– Wesley Strik
Jan 7 at 12:52
add a comment |
$begingroup$
Consider $xneq 0$. Then there is an $n_0$ such that $|x|>1/n hspace{0.3cm}forall ngeq n_0 $: thus $f_n(x)=0$ definitively, and the pointwise limit is 0.
You are left only with $x=0$, but $f_n(0)=1 hspace{0.2cm}forall n$.
answered Jan 7 at 12:35
F.BattistoniF.Battistoni
515
New contributor
F.Battistoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $xneq 0$?)
But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.
answered Jan 7 at 12:33
A. PongráczA. Pongrácz
5,9381929
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your answer is not correct. If $x=0$ then $f_n(x)=1$ for all $n$ and the limit is $1$. If $x neq 0$ then $f_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.
answered Jan 7 at 12:32
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
$begingroup$
yes that makes sense
$endgroup$
– Wesley Strik
Jan 7 at 12:52
add a comment |
$begingroup$
Your answer is not correct. If $x=0$ then $f_n(x)=1$ for all $n$ and the limit is $1$. If $x neq 0$ then $f_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.
answered Jan 7 at 12:32
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
$begingroup$
yes that makes sense
$endgroup$
– Wesley Strik
Jan 7 at 12:52
add a comment |
$begingroup$
Your answer is not correct. If $x=0$ then $f_n(x)=1$ for all $n$ and the limit is $1$. If $x neq 0$ then $f_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.
answered Jan 7 at 12:32
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
Your answer is not correct. If $x=0$ then $f_n(x)=1$ for all $n$ and the limit is $1$. If $x neq 0$ then $f_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.
answered Jan 7 at 12:32
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Jan 7 at 12:32
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Jan 7 at 12:32
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Jan 7 at 12:32
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
52.8k32055
$begingroup$
yes that makes sense
$endgroup$
– Wesley Strik
Jan 7 at 12:52
add a comment |
$begingroup$
yes that makes sense
$endgroup$
– Wesley Strik
Jan 7 at 12:52
$begingroup$
yes that makes sense
$endgroup$
– Wesley Strik
Jan 7 at 12:52
$begingroup$
yes that makes sense
$endgroup$
– Wesley Strik
Jan 7 at 12:52
add a comment |
$begingroup$
Consider $xneq 0$. Then there is an $n_0$ such that $|x|>1/n hspace{0.3cm}forall ngeq n_0 $: thus $f_n(x)=0$ definitively, and the pointwise limit is 0.
You are left only with $x=0$, but $f_n(0)=1 hspace{0.2cm}forall n$.
answered Jan 7 at 12:35
F.BattistoniF.Battistoni
515
New contributor
F.Battistoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Consider $xneq 0$. Then there is an $n_0$ such that $|x|>1/n hspace{0.3cm}forall ngeq n_0 $: thus $f_n(x)=0$ definitively, and the pointwise limit is 0.
You are left only with $x=0$, but $f_n(0)=1 hspace{0.2cm}forall n$.
answered Jan 7 at 12:35
F.BattistoniF.Battistoni
515
New contributor
F.Battistoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Consider $xneq 0$. Then there is an $n_0$ such that $|x|>1/n hspace{0.3cm}forall ngeq n_0 $: thus $f_n(x)=0$ definitively, and the pointwise limit is 0.
You are left only with $x=0$, but $f_n(0)=1 hspace{0.2cm}forall n$.
answered Jan 7 at 12:35
F.BattistoniF.Battistoni
515
New contributor
F.Battistoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Consider $xneq 0$. Then there is an $n_0$ such that $|x|>1/n hspace{0.3cm}forall ngeq n_0 $: thus $f_n(x)=0$ definitively, and the pointwise limit is 0.
You are left only with $x=0$, but $f_n(0)=1 hspace{0.2cm}forall n$.
answered Jan 7 at 12:35
F.BattistoniF.Battistoni
515
New contributor
F.Battistoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 7 at 12:35
F.BattistoniF.Battistoni
515
New contributor
F.Battistoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 7 at 12:35
F.BattistoniF.Battistoni
515
answered Jan 7 at 12:35
F.BattistoniF.Battistoni
515
515
New contributor
F.Battistoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
F.Battistoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
F.Battistoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $xneq 0$?)
But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.
answered Jan 7 at 12:33
A. PongráczA. Pongrácz
5,9381929
$endgroup$
add a comment |
$begingroup$
Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $xneq 0$?)
But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.
answered Jan 7 at 12:33
A. PongráczA. Pongrácz
5,9381929
$endgroup$
add a comment |
$begingroup$
Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $xneq 0$?)
But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.
answered Jan 7 at 12:33
A. PongráczA. Pongrácz
5,9381929
$endgroup$
Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $xneq 0$?)
But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.
answered Jan 7 at 12:33
A. PongráczA. Pongrácz
5,9381929
edited Jan 7 at 12:59
answered Jan 7 at 12:33
A. PongráczA. Pongrácz
5,9381929
answered Jan 7 at 12:33
A. PongráczA. Pongrácz
5,9381929
answered Jan 7 at 12:33
A. PongráczA. Pongrácz
5,9381929
5,9381929
add a comment |
add a comment |
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