If $f$ is continuous $displaystylelim_{x to infty}f(x)=infty$ iff $f^{-1}(K)$ is compact for every compact $K...
$begingroup$
Let $f: mathbb{R}^{m}to mathbb{R}^{n}$ be a continuous function. Prove that the following statements are equivalents:
(i) $displaystylelim_{x to infty}f(x)=infty$;
(ii) $f^{-1}(K)$ is compact for every compact $K subset mathbb{R}^{n}$.
My attempt.
(i) $Longrightarrow$ (ii): If $K subset mathbb{R}^{n}$ is compact, $f^{-1}(K)$ is closed, since $f$ is continuous. If $f^{-1}(K)$ is unbounded, there is a sequence $(x_{n}) subset f^{-1}(K)$ such that $x_{n} to infty$.
I want to conclude that $f(x_{n}) to infty$. How can I ensure it?
(ii) $Longrightarrow$ (i): I dont have a good idea. Can someone help me? I want just a hint, not a solution.
real-analysis continuity compactness
asked Jan 6 at 3:49
Lucas CorrêaLucas Corrêa
1,5351421
$endgroup$
|
show 4 more comments
$begingroup$
Let $f: mathbb{R}^{m}to mathbb{R}^{n}$ be a continuous function. Prove that the following statements are equivalents:
(i) $displaystylelim_{x to infty}f(x)=infty$;
(ii) $f^{-1}(K)$ is compact for every compact $K subset mathbb{R}^{n}$.
My attempt.
(i) $Longrightarrow$ (ii): If $K subset mathbb{R}^{n}$ is compact, $f^{-1}(K)$ is closed, since $f$ is continuous. If $f^{-1}(K)$ is unbounded, there is a sequence $(x_{n}) subset f^{-1}(K)$ such that $x_{n} to infty$.
I want to conclude that $f(x_{n}) to infty$. How can I ensure it?
(ii) $Longrightarrow$ (i): I dont have a good idea. Can someone help me? I want just a hint, not a solution.
real-analysis continuity compactness
asked Jan 6 at 3:49
Lucas CorrêaLucas Corrêa
1,5351421
$endgroup$
$begingroup$
How do you define $xtoinfty$ exactly? With the one-point compactification?
$endgroup$
– SmileyCraft
Jan 6 at 3:55
1
$begingroup$
Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
$endgroup$
– ItsJustLogicBro
Jan 6 at 3:58
$begingroup$
@SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
$endgroup$
– Lucas Corrêa
Jan 6 at 4:02
$begingroup$
@ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
$endgroup$
– mathcounterexamples.net
Jan 6 at 4:02
1
$begingroup$
@LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
$endgroup$
– ItsJustLogicBro
Jan 6 at 4:06
|
show 4 more comments
$begingroup$
Let $f: mathbb{R}^{m}to mathbb{R}^{n}$ be a continuous function. Prove that the following statements are equivalents:
(i) $displaystylelim_{x to infty}f(x)=infty$;
(ii) $f^{-1}(K)$ is compact for every compact $K subset mathbb{R}^{n}$.
My attempt.
(i) $Longrightarrow$ (ii): If $K subset mathbb{R}^{n}$ is compact, $f^{-1}(K)$ is closed, since $f$ is continuous. If $f^{-1}(K)$ is unbounded, there is a sequence $(x_{n}) subset f^{-1}(K)$ such that $x_{n} to infty$.
I want to conclude that $f(x_{n}) to infty$. How can I ensure it?
(ii) $Longrightarrow$ (i): I dont have a good idea. Can someone help me? I want just a hint, not a solution.
real-analysis continuity compactness
asked Jan 6 at 3:49
Lucas CorrêaLucas Corrêa
1,5351421
$endgroup$
Let $f: mathbb{R}^{m}to mathbb{R}^{n}$ be a continuous function. Prove that the following statements are equivalents:
(i) $displaystylelim_{x to infty}f(x)=infty$;
(ii) $f^{-1}(K)$ is compact for every compact $K subset mathbb{R}^{n}$.
My attempt.
(i) $Longrightarrow$ (ii): If $K subset mathbb{R}^{n}$ is compact, $f^{-1}(K)$ is closed, since $f$ is continuous. If $f^{-1}(K)$ is unbounded, there is a sequence $(x_{n}) subset f^{-1}(K)$ such that $x_{n} to infty$.
I want to conclude that $f(x_{n}) to infty$. How can I ensure it?
(ii) $Longrightarrow$ (i): I dont have a good idea. Can someone help me? I want just a hint, not a solution.
real-analysis continuity compactness
real-analysis continuity compactness
asked Jan 6 at 3:49
Lucas CorrêaLucas Corrêa
1,5351421
asked Jan 6 at 3:49
Lucas CorrêaLucas Corrêa
1,5351421
asked Jan 6 at 3:49
Lucas CorrêaLucas Corrêa
1,5351421
asked Jan 6 at 3:49
Lucas CorrêaLucas Corrêa
1,5351421
asked Jan 6 at 3:49
Lucas CorrêaLucas Corrêa
1,5351421
1,5351421
$begingroup$
How do you define $xtoinfty$ exactly? With the one-point compactification?
$endgroup$
– SmileyCraft
Jan 6 at 3:55
1
$begingroup$
Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
$endgroup$
– ItsJustLogicBro
Jan 6 at 3:58
$begingroup$
@SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
$endgroup$
– Lucas Corrêa
Jan 6 at 4:02
$begingroup$
@ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
$endgroup$
– mathcounterexamples.net
Jan 6 at 4:02
1
$begingroup$
@LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
$endgroup$
– ItsJustLogicBro
Jan 6 at 4:06
|
show 4 more comments
$begingroup$
How do you define $xtoinfty$ exactly? With the one-point compactification?
$endgroup$
– SmileyCraft
Jan 6 at 3:55
1
$begingroup$
Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
$endgroup$
– ItsJustLogicBro
Jan 6 at 3:58
$begingroup$
@SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
$endgroup$
– Lucas Corrêa
Jan 6 at 4:02
$begingroup$
@ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
$endgroup$
– mathcounterexamples.net
Jan 6 at 4:02
1
$begingroup$
@LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
$endgroup$
– ItsJustLogicBro
Jan 6 at 4:06
$begingroup$
How do you define $xtoinfty$ exactly? With the one-point compactification?
$endgroup$
– SmileyCraft
Jan 6 at 3:55
$begingroup$
How do you define $xtoinfty$ exactly? With the one-point compactification?
$endgroup$
– SmileyCraft
Jan 6 at 3:55
1
1
$begingroup$
Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
$endgroup$
– ItsJustLogicBro
Jan 6 at 3:58
$begingroup$
Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
$endgroup$
– ItsJustLogicBro
Jan 6 at 3:58
$begingroup$
@SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
$endgroup$
– Lucas Corrêa
Jan 6 at 4:02
$begingroup$
@SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
$endgroup$
– Lucas Corrêa
Jan 6 at 4:02
$begingroup$
@ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
$endgroup$
– mathcounterexamples.net
Jan 6 at 4:02
$begingroup$
@ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
$endgroup$
– mathcounterexamples.net
Jan 6 at 4:02
1
1
$begingroup$
@LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
$endgroup$
– ItsJustLogicBro
Jan 6 at 4:06
$begingroup$
@LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
$endgroup$
– ItsJustLogicBro
Jan 6 at 4:06
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
To complete your argument for $(i)$ $|x_n|rightarrow infty$ implies that $f(x_n)rightarrow infty$ by hypothesis.
ii) implies i)Suppose that there exists a sequence $x_n$ such that $lim_n|x_n|=infty$ and $f(x_n)$ is bounded, $f(x_n)$ is contained in a closed ball $B$ which is compact, so $f^{-1}(B)=C$ is compact, and bounded by hypothesis, but $x_nin C$ contradiction since $x_n$ is not bounded.
answered Jan 6 at 3:59
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
add a comment |
$begingroup$
The definition of $x_ntoinfty$ I am used to is that for every compact set $K$ there exists $N$ such that for all $n>N$ we have $x_nnotin K$. So then the statement $$lim_{xtoinfty}f(x)=infty$$ would literally mean that for every compact set $K$ there exists compact set $L$ such that for all $xnotin L$ we have $f(x)notin K$, so $f^{-1}(K)subseteq L$.
I would like to mention how this elegant definition can be made precise with topology. In topology we say that $x_nto x$ if for every open set $U$ containing $x$ there exists $N$ such that for all $n>N$ we have $x_nin U$. There is also a thing in topology called the one-point compactification of a topological space $X$. The idea is to add a "point at infinity", so you get $Xcup{infty}$, and the open sets containing $infty$ are exactly the complements of the compact sets in $X$.
answered Jan 6 at 4:18
SmileyCraftSmileyCraft
3,749519
$endgroup$
$begingroup$
Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
$endgroup$
– Lucas Corrêa
Jan 6 at 12:58
add a comment |
$begingroup$
Regarding (ii) $Longrightarrow$ (i)
Take $p in mathbb N$. You want to find $M>0$ such that for $Vert x Vert >M$ you have $Vert f(x) Vert >p$.
The closed ball $K_p$centered on zero with radius $p$ is a compact subset of $mathbb R^n$. $f^{-1}(mathbb R^n setminus K_p) = f^{-1}(mathbb R^n)setminus f^{-1}(K_p)$. And $f^{-1}(K_p)$ is by hypothesis a compact subset of $mathbb R^n$, therefore bounded and included in a closed ball centered on zero with a radius $M>0$.
Where are done: the image of a point $x$ with a norm $>M$ is such that $Vert f(x)Vert >p$.
answered Jan 6 at 4:12
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
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$begingroup$
To complete your argument for $(i)$ $|x_n|rightarrow infty$ implies that $f(x_n)rightarrow infty$ by hypothesis.
ii) implies i)Suppose that there exists a sequence $x_n$ such that $lim_n|x_n|=infty$ and $f(x_n)$ is bounded, $f(x_n)$ is contained in a closed ball $B$ which is compact, so $f^{-1}(B)=C$ is compact, and bounded by hypothesis, but $x_nin C$ contradiction since $x_n$ is not bounded.
answered Jan 6 at 3:59
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
add a comment |
$begingroup$
To complete your argument for $(i)$ $|x_n|rightarrow infty$ implies that $f(x_n)rightarrow infty$ by hypothesis.
ii) implies i)Suppose that there exists a sequence $x_n$ such that $lim_n|x_n|=infty$ and $f(x_n)$ is bounded, $f(x_n)$ is contained in a closed ball $B$ which is compact, so $f^{-1}(B)=C$ is compact, and bounded by hypothesis, but $x_nin C$ contradiction since $x_n$ is not bounded.
answered Jan 6 at 3:59
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
add a comment |
$begingroup$
To complete your argument for $(i)$ $|x_n|rightarrow infty$ implies that $f(x_n)rightarrow infty$ by hypothesis.
ii) implies i)Suppose that there exists a sequence $x_n$ such that $lim_n|x_n|=infty$ and $f(x_n)$ is bounded, $f(x_n)$ is contained in a closed ball $B$ which is compact, so $f^{-1}(B)=C$ is compact, and bounded by hypothesis, but $x_nin C$ contradiction since $x_n$ is not bounded.
answered Jan 6 at 3:59
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
To complete your argument for $(i)$ $|x_n|rightarrow infty$ implies that $f(x_n)rightarrow infty$ by hypothesis.
ii) implies i)Suppose that there exists a sequence $x_n$ such that $lim_n|x_n|=infty$ and $f(x_n)$ is bounded, $f(x_n)$ is contained in a closed ball $B$ which is compact, so $f^{-1}(B)=C$ is compact, and bounded by hypothesis, but $x_nin C$ contradiction since $x_n$ is not bounded.
answered Jan 6 at 3:59
Tsemo AristideTsemo Aristide
59.6k11446
edited Jan 6 at 4:05
answered Jan 6 at 3:59
Tsemo AristideTsemo Aristide
59.6k11446
answered Jan 6 at 3:59
Tsemo AristideTsemo Aristide
59.6k11446
answered Jan 6 at 3:59
Tsemo AristideTsemo Aristide
59.6k11446
59.6k11446
add a comment |
add a comment |
$begingroup$
The definition of $x_ntoinfty$ I am used to is that for every compact set $K$ there exists $N$ such that for all $n>N$ we have $x_nnotin K$. So then the statement $$lim_{xtoinfty}f(x)=infty$$ would literally mean that for every compact set $K$ there exists compact set $L$ such that for all $xnotin L$ we have $f(x)notin K$, so $f^{-1}(K)subseteq L$.
I would like to mention how this elegant definition can be made precise with topology. In topology we say that $x_nto x$ if for every open set $U$ containing $x$ there exists $N$ such that for all $n>N$ we have $x_nin U$. There is also a thing in topology called the one-point compactification of a topological space $X$. The idea is to add a "point at infinity", so you get $Xcup{infty}$, and the open sets containing $infty$ are exactly the complements of the compact sets in $X$.
answered Jan 6 at 4:18
SmileyCraftSmileyCraft
3,749519
$endgroup$
$begingroup$
Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
$endgroup$
– Lucas Corrêa
Jan 6 at 12:58
add a comment |
$begingroup$
The definition of $x_ntoinfty$ I am used to is that for every compact set $K$ there exists $N$ such that for all $n>N$ we have $x_nnotin K$. So then the statement $$lim_{xtoinfty}f(x)=infty$$ would literally mean that for every compact set $K$ there exists compact set $L$ such that for all $xnotin L$ we have $f(x)notin K$, so $f^{-1}(K)subseteq L$.
I would like to mention how this elegant definition can be made precise with topology. In topology we say that $x_nto x$ if for every open set $U$ containing $x$ there exists $N$ such that for all $n>N$ we have $x_nin U$. There is also a thing in topology called the one-point compactification of a topological space $X$. The idea is to add a "point at infinity", so you get $Xcup{infty}$, and the open sets containing $infty$ are exactly the complements of the compact sets in $X$.
answered Jan 6 at 4:18
SmileyCraftSmileyCraft
3,749519
$endgroup$
$begingroup$
Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
$endgroup$
– Lucas Corrêa
Jan 6 at 12:58
add a comment |
$begingroup$
The definition of $x_ntoinfty$ I am used to is that for every compact set $K$ there exists $N$ such that for all $n>N$ we have $x_nnotin K$. So then the statement $$lim_{xtoinfty}f(x)=infty$$ would literally mean that for every compact set $K$ there exists compact set $L$ such that for all $xnotin L$ we have $f(x)notin K$, so $f^{-1}(K)subseteq L$.
I would like to mention how this elegant definition can be made precise with topology. In topology we say that $x_nto x$ if for every open set $U$ containing $x$ there exists $N$ such that for all $n>N$ we have $x_nin U$. There is also a thing in topology called the one-point compactification of a topological space $X$. The idea is to add a "point at infinity", so you get $Xcup{infty}$, and the open sets containing $infty$ are exactly the complements of the compact sets in $X$.
answered Jan 6 at 4:18
SmileyCraftSmileyCraft
3,749519
$endgroup$
The definition of $x_ntoinfty$ I am used to is that for every compact set $K$ there exists $N$ such that for all $n>N$ we have $x_nnotin K$. So then the statement $$lim_{xtoinfty}f(x)=infty$$ would literally mean that for every compact set $K$ there exists compact set $L$ such that for all $xnotin L$ we have $f(x)notin K$, so $f^{-1}(K)subseteq L$.
I would like to mention how this elegant definition can be made precise with topology. In topology we say that $x_nto x$ if for every open set $U$ containing $x$ there exists $N$ such that for all $n>N$ we have $x_nin U$. There is also a thing in topology called the one-point compactification of a topological space $X$. The idea is to add a "point at infinity", so you get $Xcup{infty}$, and the open sets containing $infty$ are exactly the complements of the compact sets in $X$.
answered Jan 6 at 4:18
SmileyCraftSmileyCraft
3,749519
answered Jan 6 at 4:18
SmileyCraftSmileyCraft
3,749519
answered Jan 6 at 4:18
SmileyCraftSmileyCraft
3,749519
answered Jan 6 at 4:18
SmileyCraftSmileyCraft
3,749519
3,749519
$begingroup$
Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
$endgroup$
– Lucas Corrêa
Jan 6 at 12:58
add a comment |
$begingroup$
Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
$endgroup$
– Lucas Corrêa
Jan 6 at 12:58
$begingroup$
Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
$endgroup$
– Lucas Corrêa
Jan 6 at 12:58
$begingroup$
Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
$endgroup$
– Lucas Corrêa
Jan 6 at 12:58
add a comment |
$begingroup$
Regarding (ii) $Longrightarrow$ (i)
Take $p in mathbb N$. You want to find $M>0$ such that for $Vert x Vert >M$ you have $Vert f(x) Vert >p$.
The closed ball $K_p$centered on zero with radius $p$ is a compact subset of $mathbb R^n$. $f^{-1}(mathbb R^n setminus K_p) = f^{-1}(mathbb R^n)setminus f^{-1}(K_p)$. And $f^{-1}(K_p)$ is by hypothesis a compact subset of $mathbb R^n$, therefore bounded and included in a closed ball centered on zero with a radius $M>0$.
Where are done: the image of a point $x$ with a norm $>M$ is such that $Vert f(x)Vert >p$.
answered Jan 6 at 4:12
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
Regarding (ii) $Longrightarrow$ (i)
Take $p in mathbb N$. You want to find $M>0$ such that for $Vert x Vert >M$ you have $Vert f(x) Vert >p$.
The closed ball $K_p$centered on zero with radius $p$ is a compact subset of $mathbb R^n$. $f^{-1}(mathbb R^n setminus K_p) = f^{-1}(mathbb R^n)setminus f^{-1}(K_p)$. And $f^{-1}(K_p)$ is by hypothesis a compact subset of $mathbb R^n$, therefore bounded and included in a closed ball centered on zero with a radius $M>0$.
Where are done: the image of a point $x$ with a norm $>M$ is such that $Vert f(x)Vert >p$.
answered Jan 6 at 4:12
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
Regarding (ii) $Longrightarrow$ (i)
Take $p in mathbb N$. You want to find $M>0$ such that for $Vert x Vert >M$ you have $Vert f(x) Vert >p$.
The closed ball $K_p$centered on zero with radius $p$ is a compact subset of $mathbb R^n$. $f^{-1}(mathbb R^n setminus K_p) = f^{-1}(mathbb R^n)setminus f^{-1}(K_p)$. And $f^{-1}(K_p)$ is by hypothesis a compact subset of $mathbb R^n$, therefore bounded and included in a closed ball centered on zero with a radius $M>0$.
Where are done: the image of a point $x$ with a norm $>M$ is such that $Vert f(x)Vert >p$.
answered Jan 6 at 4:12
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
Regarding (ii) $Longrightarrow$ (i)
Take $p in mathbb N$. You want to find $M>0$ such that for $Vert x Vert >M$ you have $Vert f(x) Vert >p$.
The closed ball $K_p$centered on zero with radius $p$ is a compact subset of $mathbb R^n$. $f^{-1}(mathbb R^n setminus K_p) = f^{-1}(mathbb R^n)setminus f^{-1}(K_p)$. And $f^{-1}(K_p)$ is by hypothesis a compact subset of $mathbb R^n$, therefore bounded and included in a closed ball centered on zero with a radius $M>0$.
Where are done: the image of a point $x$ with a norm $>M$ is such that $Vert f(x)Vert >p$.
answered Jan 6 at 4:12
mathcounterexamples.netmathcounterexamples.net
27k22158
edited Jan 6 at 9:09
answered Jan 6 at 4:12
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 6 at 4:12
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 6 at 4:12
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
add a comment |
add a comment |
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How do you define $xtoinfty$ exactly? With the one-point compactification?
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– SmileyCraft
Jan 6 at 3:55
1
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Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
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– ItsJustLogicBro
Jan 6 at 3:58
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@SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
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– Lucas Corrêa
Jan 6 at 4:02
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@ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
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– mathcounterexamples.net
Jan 6 at 4:02
1
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@LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
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– ItsJustLogicBro
Jan 6 at 4:06