Doubt about definition of infinite limit and limit as x tends to infinity
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In the limits and continuity chapter in my textbook, the following definitions are given:
(1) For limit of f(x) as x tends to infinity:
We say that $ÃÂ(x)$ has the limit $L$ as $x$ approaches infinity and write $lim_xto infty f(x) = L$ if, for every number $epsilon > 0$ there exists a corresponding number $M>0$ such that for all $x>M, |f(x) - L| < epsilon$.
(2) For infinite limits:
We say that $ÃÂ(x)$ approaches infinity as $x$ approaches $x_0$, and write $lim_xto x_0 ÃÂ(x) = infty$, if for every number $B>0$ there exists a corresponding $delta > 0$ such that for all $x$,
$$0 < |x - x_0| < delta implies f(x) > B$$
What I don't understand is why the numbers M and B have to be greater than zero in the first and second definitions respectively. I think the definition would work without this added constraint, i.e. if M (and, similarly, B) is any real number.
So, I don't understand why this condition is present in the definitions.
Since I couldn't find any discussion on this topic anywhere on the net or in my book, I have asked this question.
What am I missing?
definition epsilon-delta constraints
add a comment |Â
up vote
0
down vote
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In the limits and continuity chapter in my textbook, the following definitions are given:
(1) For limit of f(x) as x tends to infinity:
We say that $ÃÂ(x)$ has the limit $L$ as $x$ approaches infinity and write $lim_xto infty f(x) = L$ if, for every number $epsilon > 0$ there exists a corresponding number $M>0$ such that for all $x>M, |f(x) - L| < epsilon$.
(2) For infinite limits:
We say that $ÃÂ(x)$ approaches infinity as $x$ approaches $x_0$, and write $lim_xto x_0 ÃÂ(x) = infty$, if for every number $B>0$ there exists a corresponding $delta > 0$ such that for all $x$,
$$0 < |x - x_0| < delta implies f(x) > B$$
What I don't understand is why the numbers M and B have to be greater than zero in the first and second definitions respectively. I think the definition would work without this added constraint, i.e. if M (and, similarly, B) is any real number.
So, I don't understand why this condition is present in the definitions.
Since I couldn't find any discussion on this topic anywhere on the net or in my book, I have asked this question.
What am I missing?
definition epsilon-delta constraints
3
You're right. The demands of $M$ and $B$ are positive are superfluous.
â MathOverview
Sep 9 at 13:25
@MathOverview, but then why are they included in the definitions?
â Mr Reality
Sep 9 at 13:30
I think it's just to stay consistent with the other limite definitions.
â MathOverview
Sep 9 at 13:35
@MathOverview You should write that as an answer.
â Noah Schweber
Sep 9 at 14:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In the limits and continuity chapter in my textbook, the following definitions are given:
(1) For limit of f(x) as x tends to infinity:
We say that $ÃÂ(x)$ has the limit $L$ as $x$ approaches infinity and write $lim_xto infty f(x) = L$ if, for every number $epsilon > 0$ there exists a corresponding number $M>0$ such that for all $x>M, |f(x) - L| < epsilon$.
(2) For infinite limits:
We say that $ÃÂ(x)$ approaches infinity as $x$ approaches $x_0$, and write $lim_xto x_0 ÃÂ(x) = infty$, if for every number $B>0$ there exists a corresponding $delta > 0$ such that for all $x$,
$$0 < |x - x_0| < delta implies f(x) > B$$
What I don't understand is why the numbers M and B have to be greater than zero in the first and second definitions respectively. I think the definition would work without this added constraint, i.e. if M (and, similarly, B) is any real number.
So, I don't understand why this condition is present in the definitions.
Since I couldn't find any discussion on this topic anywhere on the net or in my book, I have asked this question.
What am I missing?
definition epsilon-delta constraints
In the limits and continuity chapter in my textbook, the following definitions are given:
(1) For limit of f(x) as x tends to infinity:
We say that $ÃÂ(x)$ has the limit $L$ as $x$ approaches infinity and write $lim_xto infty f(x) = L$ if, for every number $epsilon > 0$ there exists a corresponding number $M>0$ such that for all $x>M, |f(x) - L| < epsilon$.
(2) For infinite limits:
We say that $ÃÂ(x)$ approaches infinity as $x$ approaches $x_0$, and write $lim_xto x_0 ÃÂ(x) = infty$, if for every number $B>0$ there exists a corresponding $delta > 0$ such that for all $x$,
$$0 < |x - x_0| < delta implies f(x) > B$$
What I don't understand is why the numbers M and B have to be greater than zero in the first and second definitions respectively. I think the definition would work without this added constraint, i.e. if M (and, similarly, B) is any real number.
So, I don't understand why this condition is present in the definitions.
Since I couldn't find any discussion on this topic anywhere on the net or in my book, I have asked this question.
What am I missing?
definition epsilon-delta constraints
definition epsilon-delta constraints
asked Sep 9 at 13:18
Mr Reality
1469
1469
3
You're right. The demands of $M$ and $B$ are positive are superfluous.
â MathOverview
Sep 9 at 13:25
@MathOverview, but then why are they included in the definitions?
â Mr Reality
Sep 9 at 13:30
I think it's just to stay consistent with the other limite definitions.
â MathOverview
Sep 9 at 13:35
@MathOverview You should write that as an answer.
â Noah Schweber
Sep 9 at 14:04
add a comment |Â
3
You're right. The demands of $M$ and $B$ are positive are superfluous.
â MathOverview
Sep 9 at 13:25
@MathOverview, but then why are they included in the definitions?
â Mr Reality
Sep 9 at 13:30
I think it's just to stay consistent with the other limite definitions.
â MathOverview
Sep 9 at 13:35
@MathOverview You should write that as an answer.
â Noah Schweber
Sep 9 at 14:04
3
3
You're right. The demands of $M$ and $B$ are positive are superfluous.
â MathOverview
Sep 9 at 13:25
You're right. The demands of $M$ and $B$ are positive are superfluous.
â MathOverview
Sep 9 at 13:25
@MathOverview, but then why are they included in the definitions?
â Mr Reality
Sep 9 at 13:30
@MathOverview, but then why are they included in the definitions?
â Mr Reality
Sep 9 at 13:30
I think it's just to stay consistent with the other limite definitions.
â MathOverview
Sep 9 at 13:35
I think it's just to stay consistent with the other limite definitions.
â MathOverview
Sep 9 at 13:35
@MathOverview You should write that as an answer.
â Noah Schweber
Sep 9 at 14:04
@MathOverview You should write that as an answer.
â Noah Schweber
Sep 9 at 14:04
add a comment |Â
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3
You're right. The demands of $M$ and $B$ are positive are superfluous.
â MathOverview
Sep 9 at 13:25
@MathOverview, but then why are they included in the definitions?
â Mr Reality
Sep 9 at 13:30
I think it's just to stay consistent with the other limite definitions.
â MathOverview
Sep 9 at 13:35
@MathOverview You should write that as an answer.
â Noah Schweber
Sep 9 at 14:04