Limit to 0 ambiguity [on hold] The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Need help with a limitHow to evaluate indeterminate form of a limitThe limit of $ lnx + cotx$Limit of a rational function to the power of xEvaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Evaluate the following limit without L'HopitalLimit of: $ -x+sqrtx^2+x $ for $ xtoinfty $Limit with integral and powerWhat is the result of the following limit?Determining if a multivariable limit exists
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Limit to 0 ambiguity [on hold]
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Need help with a limitHow to evaluate indeterminate form of a limitThe limit of $ lnx + cotx$Limit of a rational function to the power of xEvaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Evaluate the following limit without L'HopitalLimit of: $ -x+sqrtx^2+x $ for $ xtoinfty $Limit with integral and powerWhat is the result of the following limit?Determining if a multivariable limit exists
$begingroup$
I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.
limits
edited Apr 10 at 3:15
user8718165
1167
asked Apr 10 at 1:50
J.MohJ.Moh
535
$endgroup$
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo Apr 10 at 7:26
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
add a comment |
$begingroup$
I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.
limits
edited Apr 10 at 3:15
user8718165
1167
asked Apr 10 at 1:50
J.MohJ.Moh
535
$endgroup$
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo Apr 10 at 7:26
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
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
Apr 10 at 1:52
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
Apr 10 at 1:57
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
Apr 10 at 2:10
add a comment |
$begingroup$
I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.
limits
edited Apr 10 at 3:15
user8718165
1167
asked Apr 10 at 1:50
J.MohJ.Moh
535
$endgroup$
I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.
limits
limits
edited Apr 10 at 3:15
user8718165
1167
asked Apr 10 at 1:50
J.MohJ.Moh
535
edited Apr 10 at 3:15
user8718165
1167
asked Apr 10 at 1:50
J.MohJ.Moh
535
edited Apr 10 at 3:15
user8718165
1167
edited Apr 10 at 3:15
user8718165
1167
edited Apr 10 at 3:15
user8718165
1167
1167
asked Apr 10 at 1:50
J.MohJ.Moh
535
asked Apr 10 at 1:50
J.MohJ.Moh
535
asked Apr 10 at 1:50
J.MohJ.Moh
535
535
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo Apr 10 at 7:26
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
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo Apr 10 at 7:26
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
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
Apr 10 at 1:52
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
Apr 10 at 1:57
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
Apr 10 at 2:10
add a comment |
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
Apr 10 at 1:52
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
Apr 10 at 1:57
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
Apr 10 at 2:10
1
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
Apr 10 at 1:52
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
Apr 10 at 1:52
3
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
Apr 10 at 1:57
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
Apr 10 at 1:57
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
Apr 10 at 2:10
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
Apr 10 at 2:10
add a comment |
2 Answers
2
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.
answered Apr 10 at 1:54
MPWMPW
31.2k12157
$endgroup$
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ 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:
$$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (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, $lim_xto 2g(x)=-infty$ 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.
answered Apr 10 at 2:17
Michael RybkinMichael Rybkin
4,264422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
Apr 10 at 2:47
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.
answered Apr 10 at 1:54
MPWMPW
31.2k12157
$endgroup$
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.
answered Apr 10 at 1:54
MPWMPW
31.2k12157
$endgroup$
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.
answered Apr 10 at 1:54
MPWMPW
31.2k12157
$endgroup$
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 Apr 10 at 1:54
MPWMPW
31.2k12157
answered Apr 10 at 1:54
MPWMPW
31.2k12157
answered Apr 10 at 1:54
MPWMPW
31.2k12157
answered Apr 10 at 1:54
MPWMPW
31.2k12157
31.2k12157
add a comment |
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ 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:
$$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (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, $lim_xto 2g(x)=-infty$ 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.
answered Apr 10 at 2:17
Michael RybkinMichael Rybkin
4,264422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
Apr 10 at 2:47
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ 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:
$$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (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, $lim_xto 2g(x)=-infty$ 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.
answered Apr 10 at 2:17
Michael RybkinMichael Rybkin
4,264422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
Apr 10 at 2:47
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ 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:
$$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (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, $lim_xto 2g(x)=-infty$ 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.
answered Apr 10 at 2:17
Michael RybkinMichael Rybkin
4,264422
$endgroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ 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:
$$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (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, $lim_xto 2g(x)=-infty$ 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.
answered Apr 10 at 2:17
Michael RybkinMichael Rybkin
4,264422
edited Apr 10 at 6:27
answered Apr 10 at 2:17
Michael RybkinMichael Rybkin
4,264422
answered Apr 10 at 2:17
Michael RybkinMichael Rybkin
4,264422
answered Apr 10 at 2:17
Michael RybkinMichael Rybkin
4,264422
4,264422
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
Apr 10 at 2:47
add a comment |
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
Apr 10 at 2:47
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
Apr 10 at 2:47
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
Apr 10 at 2:47
add a comment |
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
Apr 10 at 1:52
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
Apr 10 at 1:57
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
Apr 10 at 2:10