Limit to 0 ambiguity [on hold]
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I can't determine the limit of such form:
$$lim_{x to 0} frac{1}{x}, $$
$$+infty~text{or }-infty$$
I tried to get around it, no success.
limits
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put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I can't determine the limit of such form:
$$lim_{x to 0} frac{1}{x}, $$
$$+infty~text{or }-infty$$
I tried to get around it, no success.
limits
$endgroup$
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
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– Dave
2 days ago
3
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How are you defining a limit?
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– John Doe
2 days ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
2 days ago
add a comment |
$begingroup$
I can't determine the limit of such form:
$$lim_{x to 0} frac{1}{x}, $$
$$+infty~text{or }-infty$$
I tried to get around it, no success.
limits
$endgroup$
I can't determine the limit of such form:
$$lim_{x to 0} frac{1}{x}, $$
$$+infty~text{or }-infty$$
I tried to get around it, no success.
limits
limits
edited 2 days ago
user8718165
1167
1167
asked 2 days ago
J.MohJ.Moh
535
535
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
$endgroup$
– Dave
2 days ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
2 days ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
2 days ago
add a comment |
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
$endgroup$
– Dave
2 days ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
2 days ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
2 days ago
1
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
$endgroup$
– Dave
2 days ago
$begingroup$
Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
$endgroup$
– Dave
2 days ago
3
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
2 days ago
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
2 days ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
2 days ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
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The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.
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add a comment |
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It's instructive to take a look at the graph of $f(x)=frac{1}{x}$ to better see what exactly is going on with the function as $x$ goes to zero:

As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:
$$lim_{xto 0^-}frac{1}{x}=-infty, lim_{xto 0^+}frac{1}{x}=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_{xto 0^-}frac{1}{x}nelim_{xto 0^+}frac{1}{x}$. Thus, $lim_{xto 0}frac{1}{x}=DNE$ (does not exist).
Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_{xto 2}g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.
$endgroup$
add a comment |
$begingroup$
The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.
$endgroup$
add a comment |
$begingroup$
The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.
$endgroup$
The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.
answered 2 days ago
MPWMPW
31.2k12157
31.2k12157
add a comment |
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac{1}{x}$ to better see what exactly is going on with the function as $x$ goes to zero:

As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:
$$lim_{xto 0^-}frac{1}{x}=-infty, lim_{xto 0^+}frac{1}{x}=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_{xto 0^-}frac{1}{x}nelim_{xto 0^+}frac{1}{x}$. Thus, $lim_{xto 0}frac{1}{x}=DNE$ (does not exist).
Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_{xto 2}g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 days ago
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac{1}{x}$ to better see what exactly is going on with the function as $x$ goes to zero:

As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:
$$lim_{xto 0^-}frac{1}{x}=-infty, lim_{xto 0^+}frac{1}{x}=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_{xto 0^-}frac{1}{x}nelim_{xto 0^+}frac{1}{x}$. Thus, $lim_{xto 0}frac{1}{x}=DNE$ (does not exist).
Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_{xto 2}g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 days ago
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac{1}{x}$ to better see what exactly is going on with the function as $x$ goes to zero:

As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:
$$lim_{xto 0^-}frac{1}{x}=-infty, lim_{xto 0^+}frac{1}{x}=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_{xto 0^-}frac{1}{x}nelim_{xto 0^+}frac{1}{x}$. Thus, $lim_{xto 0}frac{1}{x}=DNE$ (does not exist).
Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_{xto 2}g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.
$endgroup$
It's instructive to take a look at the graph of $f(x)=frac{1}{x}$ to better see what exactly is going on with the function as $x$ goes to zero:

As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:
$$lim_{xto 0^-}frac{1}{x}=-infty, lim_{xto 0^+}frac{1}{x}=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_{xto 0^-}frac{1}{x}nelim_{xto 0^+}frac{1}{x}$. Thus, $lim_{xto 0}frac{1}{x}=DNE$ (does not exist).
Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_{xto 2}g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.
edited 2 days ago
answered 2 days ago
Michael RybkinMichael Rybkin
4,264422
4,264422
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 days ago
add a comment |
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 days ago
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 days ago
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 days ago
add a comment |
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
$endgroup$
– Dave
2 days ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
2 days ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
2 days ago