Limit to 0 ambiguity The 2019 Stack Overflow Developer Survey Results Are InLimit Calculation - Sequences at infinityNeed help with a limitStrategy for tackling $lim_ntoinftyfracn(ln n)^-p.$Evaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Finding a basic limitCan the limit of a polynomial involving infinity be finite?Limit 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
The 2019 Stack Overflow Developer Survey Results Are InLimit Calculation - Sequences at infinityNeed help with a limitStrategy for tackling $lim_ntoinftyfracn(ln n)^-p.$Evaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Finding a basic limitCan the limit of a polynomial involving infinity be finite?Limit 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 8 mins ago
user8718165
1167
asked 1 hour ago
J.MohJ.Moh
475
$endgroup$
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 8 mins ago
user8718165
1167
asked 1 hour ago
J.MohJ.Moh
475
$endgroup$
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
1 hour ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
1 hour ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
1 hour 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 8 mins ago
user8718165
1167
asked 1 hour ago
J.MohJ.Moh
475
$endgroup$
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 8 mins ago
user8718165
1167
asked 1 hour ago
J.MohJ.Moh
475
edited 8 mins ago
user8718165
1167
asked 1 hour ago
J.MohJ.Moh
475
edited 8 mins ago
user8718165
1167
edited 8 mins ago
user8718165
1167
edited 8 mins ago
user8718165
1167
1167
asked 1 hour ago
J.MohJ.Moh
475
asked 1 hour ago
J.MohJ.Moh
475
asked 1 hour ago
J.MohJ.Moh
475
475
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
1 hour ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
1 hour ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
1 hour 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
1 hour ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
1 hour ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
1 hour 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
1 hour 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
1 hour ago
3
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
1 hour ago
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
1 hour ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
1 hour ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
1 hour 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 1 hour 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 47 mins ago
J. W. Tanner
4,6441320
answered 1 hour ago
Michael RybkinMichael Rybkin
4,259422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
36 mins 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
MPWMPW
31.2k12157
answered 1 hour ago
MPWMPW
31.2k12157
answered 1 hour ago
MPWMPW
31.2k12157
answered 1 hour 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 47 mins ago
J. W. Tanner
4,6441320
answered 1 hour ago
Michael RybkinMichael Rybkin
4,259422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
36 mins 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 47 mins ago
J. W. Tanner
4,6441320
answered 1 hour ago
Michael RybkinMichael Rybkin
4,259422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
36 mins 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 47 mins ago
J. W. Tanner
4,6441320
answered 1 hour ago
Michael RybkinMichael Rybkin
4,259422
$endgroup$
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 47 mins ago
J. W. Tanner
4,6441320
answered 1 hour ago
Michael RybkinMichael Rybkin
4,259422
edited 47 mins ago
J. W. Tanner
4,6441320
edited 47 mins ago
J. W. Tanner
4,6441320
edited 47 mins ago
J. W. Tanner
4,6441320
4,6441320
answered 1 hour ago
Michael RybkinMichael Rybkin
4,259422
answered 1 hour ago
Michael RybkinMichael Rybkin
4,259422
answered 1 hour ago
Michael RybkinMichael Rybkin
4,259422
4,259422
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
36 mins ago
add a comment |
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
36 mins ago
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
36 mins ago
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
36 mins ago
add a comment |
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zHEJaJxGeh9NV2klSkzHugb7wVSe,ip4,6n4ptXdBiboxE Z,tgw8Vow IB oAe,wFmF3XYKrScbad81Dgf,3tNGkp
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
1 hour ago
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
1 hour ago
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
1 hour ago