Limit to 0 ambiguity The 2019 Stack Overflow Developer Survey Results Are InNeed 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
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Limit to 0 ambiguity
The 2019 Stack Overflow Developer Survey Results Are InNeed 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 2 hours ago
user8718165
1167
asked 3 hours ago
J.MohJ.Moh
505
$endgroup$
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 2 hours ago
user8718165
1167
asked 3 hours ago
J.MohJ.Moh
505
$endgroup$
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
3 hours ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
3 hours ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
3 hours ago
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 2 hours ago
user8718165
1167
asked 3 hours ago
J.MohJ.Moh
505
$endgroup$
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 2 hours ago
user8718165
1167
asked 3 hours ago
J.MohJ.Moh
505
edited 2 hours ago
user8718165
1167
asked 3 hours ago
J.MohJ.Moh
505
edited 2 hours ago
user8718165
1167
edited 2 hours ago
user8718165
1167
edited 2 hours ago
user8718165
1167
1167
asked 3 hours ago
J.MohJ.Moh
505
asked 3 hours ago
J.MohJ.Moh
505
asked 3 hours ago
J.MohJ.Moh
505
505
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
3 hours ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
3 hours ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
3 hours ago
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
3 hours ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
3 hours ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
3 hours ago
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
3 hours ago
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
3 hours ago
3
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
3 hours ago
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
3 hours ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
3 hours ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
3 hours ago
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 3 hours ago
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 as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is 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.
edited 2 hours ago
J. W. Tanner
4,6441320
answered 3 hours ago
Michael RybkinMichael Rybkin
4,269422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 hours ago
add a comment |
Your Answer
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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 3 hours ago
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 3 hours ago
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 3 hours ago
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 3 hours ago
MPWMPW
31.2k12157
answered 3 hours ago
MPWMPW
31.2k12157
answered 3 hours ago
MPWMPW
31.2k12157
answered 3 hours ago
MPWMPW
31.2k12157
31.2k12157
add a comment |
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ 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 as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is 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.
edited 2 hours ago
J. W. Tanner
4,6441320
answered 3 hours ago
Michael RybkinMichael Rybkin
4,269422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 hours ago
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 as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is 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.
edited 2 hours ago
J. W. Tanner
4,6441320
answered 3 hours ago
Michael RybkinMichael Rybkin
4,269422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 hours ago
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 as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is 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.
edited 2 hours ago
J. W. Tanner
4,6441320
answered 3 hours ago
Michael RybkinMichael Rybkin
4,269422
$endgroup$
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 as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is 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.
edited 2 hours ago
J. W. Tanner
4,6441320
answered 3 hours ago
Michael RybkinMichael Rybkin
4,269422
edited 2 hours ago
J. W. Tanner
4,6441320
edited 2 hours ago
J. W. Tanner
4,6441320
edited 2 hours ago
J. W. Tanner
4,6441320
4,6441320
answered 3 hours ago
Michael RybkinMichael Rybkin
4,269422
answered 3 hours ago
Michael RybkinMichael Rybkin
4,269422
answered 3 hours ago
Michael RybkinMichael Rybkin
4,269422
4,269422
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 hours ago
add a comment |
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 hours ago
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 hours ago
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
2 hours ago
add a comment |
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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
3 hours ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
3 hours ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
3 hours ago