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What is the limit of zero times x, as x approaches infinity?
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I am having difficulty determining is the solution for the following problem:
$$displaystyle lim_x rightarrow inftyleft( x times 0 right)$$
To clarify, this question assumes $0$ is a constant and is absolutely zero ("true zero"), and not another figure approaching or is approximately zero ("near zero").
I know that $infty *0$ is undefined, however my difficulty is that I'm unsure whether the answer to the problem is undefined because $infty *0$ is undefined.
From my understanding, a limit does not ever 'reach' infinity - it only approaches infinity, thus there are a rational amount of numbers. As $x*0=0$, when x is not $infty$, it seems to me that in all cases of x approaching infinity the answer could also be $0$.
limits
asked 2 hours ago
Roost1513
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up vote
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I am having difficulty determining is the solution for the following problem:
$$displaystyle lim_x rightarrow inftyleft( x times 0 right)$$
To clarify, this question assumes $0$ is a constant and is absolutely zero ("true zero"), and not another figure approaching or is approximately zero ("near zero").
I know that $infty *0$ is undefined, however my difficulty is that I'm unsure whether the answer to the problem is undefined because $infty *0$ is undefined.
From my understanding, a limit does not ever 'reach' infinity - it only approaches infinity, thus there are a rational amount of numbers. As $x*0=0$, when x is not $infty$, it seems to me that in all cases of x approaching infinity the answer could also be $0$.
limits
asked 2 hours ago
Roost1513
132
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Roost1513 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
Would need to evaluate the quantity in the bracket first, the way this is written. Then everything is identically zero.
– AnyAD
2 hours ago
Obviously can't apply limit laws to the product.
– AnyAD
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am having difficulty determining is the solution for the following problem:
$$displaystyle lim_x rightarrow inftyleft( x times 0 right)$$
To clarify, this question assumes $0$ is a constant and is absolutely zero ("true zero"), and not another figure approaching or is approximately zero ("near zero").
I know that $infty *0$ is undefined, however my difficulty is that I'm unsure whether the answer to the problem is undefined because $infty *0$ is undefined.
From my understanding, a limit does not ever 'reach' infinity - it only approaches infinity, thus there are a rational amount of numbers. As $x*0=0$, when x is not $infty$, it seems to me that in all cases of x approaching infinity the answer could also be $0$.
limits
asked 2 hours ago
Roost1513
132
New contributor
Roost1513 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am having difficulty determining is the solution for the following problem:
$$displaystyle lim_x rightarrow inftyleft( x times 0 right)$$
To clarify, this question assumes $0$ is a constant and is absolutely zero ("true zero"), and not another figure approaching or is approximately zero ("near zero").
I know that $infty *0$ is undefined, however my difficulty is that I'm unsure whether the answer to the problem is undefined because $infty *0$ is undefined.
From my understanding, a limit does not ever 'reach' infinity - it only approaches infinity, thus there are a rational amount of numbers. As $x*0=0$, when x is not $infty$, it seems to me that in all cases of x approaching infinity the answer could also be $0$.
limits
limits
asked 2 hours ago
Roost1513
132
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Roost1513 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Roost1513
132
New contributor
Roost1513 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Roost1513
132
New contributor
Roost1513 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Roost1513
132
asked 2 hours ago
Roost1513
132
132
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Roost1513 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Roost1513 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
Would need to evaluate the quantity in the bracket first, the way this is written. Then everything is identically zero.
– AnyAD
2 hours ago
Obviously can't apply limit laws to the product.
– AnyAD
2 hours ago
add a comment |Â
1
Would need to evaluate the quantity in the bracket first, the way this is written. Then everything is identically zero.
– AnyAD
2 hours ago
Obviously can't apply limit laws to the product.
– AnyAD
2 hours ago
1
1
Would need to evaluate the quantity in the bracket first, the way this is written. Then everything is identically zero.
– AnyAD
2 hours ago
Would need to evaluate the quantity in the bracket first, the way this is written. Then everything is identically zero.
– AnyAD
2 hours ago
Obviously can't apply limit laws to the product.
– AnyAD
2 hours ago
Obviously can't apply limit laws to the product.
– AnyAD
2 hours ago
add a comment |Â
6 Answers
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up vote
4
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accepted
Note that for any $x$ we have $xcdot 0=0$ and therefore
$$lim_xtoinfty (xcdot 0) =lim_xtoinfty 0=0$$
answered 2 hours ago
gimusi
78.4k73889
Would this change if I didn't use parenthesis?
– Roost1513
1 hour ago
@Roost1513 I've used the parenthesis just to indicate that we are taking the limit for $(xcdot 0)$ which is the identically zero function $f(x)=0$. For that reason the limit is equal to $0$.
– gimusi
1 hour ago
Thank you for the clarification, I just wanted to be absolutely certain.
– Roost1513
1 hour ago
@Roost1513 You are welcome! Bye
– gimusi
1 hour ago
@Roost1513 A similar case to be considered is $lim_xto 0 frac x x$. In that case we have an indeterminate form $0/0$ at $x=0$ but for $xneq 0$ the function $f(x)=x/x=1$. Can you conclude what the limit is?
– gimusi
1 hour ago
 |Â
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up vote
2
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For every $xin mathbf R$ it holds $0 cdot x = 0$. Using this definition, we let $f(x):= 0 cdot x = 0$ for every $xin mathbf R$. Hence, we have
$$ lim_xto +infty xcdot 0 = lim_xto +infty f(x) = lim_xto +infty 0 =0$$
from the definition of the limit! There is no need to think about something like $0 cdot (+infty)$.
answered 2 hours ago
Niklas
2,093719
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The limit is zero.
The reason is as follows. The function you are considering is $f(x) = x times 0$. But this means that $f(x)=0$ for all real $x$. The limit of this function as x tends to infinity is 0, even though as you point out $0timesinfty$ is undefined (but we do not need to calculate that here).
Formally, to show that this limit is zero, we need to show that for all $epsilon>0$ there exists a real $N$ so that $|f(x)-0|<epsilon$ for all $xge N$. But this is trivially true for any real $N$ and any $epsilon>0$.
answered 2 hours ago
Stefanie
11
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Let $f(x)=0times x$.
Then, the following statement is true:
For every $epsilon > 0$, there exists some $Min mathbb R$ such that, for all $x>M$, $|f(x) - 0| < epsilon$.
Therefore, by the definition of a limit, we can conclude that
$$lim_xtoinfty f(x) = 0$$
If you want the proof of the statement in yellow above:
Let $epsilon > 0$ be arbitrary. Then, let $M=1$. Let $x>M$ be arbitrary. Then, $|f(x)-0| = |0times x - 0| = |0-0|=0<epsilon$.
answered 2 hours ago
5xum
85.8k388154
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As others have said, $lim_xto infty 0 times x = lim_xtoinfty 0 = 0$. I'm going to expand a bit more on "$0 times infty$ is undefined".
We can't do operations with $infty$ directly, as you know. But we can do operations with "functions with limit $infty$", and if they behave well enough then that might give us reasonable definitions of things like "$0 times infty$".
However, if we replace $infty$ by "functions with limit $infty$" then we ought to do the same with $0$, i.e. replace $0$ by "functions with limit $0$". This is a reasonable thing to do, because it works for real numbers:
For any functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x) = a$ and $lim_x to infty g(x) = b$, where $a$ and $b$ are real numbers (i.e. finite), it's the case that $lim_x to infty (f(x)g(x)) = ab$.
In fact, it's better than that; if $a > 0$ is real then it's reasonable to say "$a times infty = infty$":
For any functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x) = a$, where $a$ is a real number and $a > 0$, and such that $lim_x to infty g(x) = infty$, then $lim_x to infty(f(x)g(x)) = infty$ also.
However, if $f(x)$ and $g(x)$ are such that $lim_x to infty f(x) = 0$ and $lim _x to infty g(x) = infty$, then we don't know anything about $lim_x to infty f(x)g(x)$. If $g(x) = x$, then taking $f(x) = 0$, $f(x) = a/x$ (where $a > 0$), and $f(x) = 1/sqrt x$, gives $lim_x to infty (f(x)g(x))$ to be, respectively, $0$, $a$ and $infty$. This is why we say "$0 times infty$" is undefined.
answered 1 hour ago
Christopher
5,59911527
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Answer to the first limit is 0. The answer is 0 because we apply the L-hospital rule and differentiate the function.
answered 2 hours ago
pinkhelium9
51
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1
L’Hopital rule?
– gimusi
2 hours ago
1
You don't need the L'Hospital rule, and in fact, its use just obfuscates the solution.
– 5xum
2 hours ago
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Note that for any $x$ we have $xcdot 0=0$ and therefore
$$lim_xtoinfty (xcdot 0) =lim_xtoinfty 0=0$$
answered 2 hours ago
gimusi
78.4k73889
Would this change if I didn't use parenthesis?
– Roost1513
1 hour ago
@Roost1513 I've used the parenthesis just to indicate that we are taking the limit for $(xcdot 0)$ which is the identically zero function $f(x)=0$. For that reason the limit is equal to $0$.
– gimusi
1 hour ago
Thank you for the clarification, I just wanted to be absolutely certain.
– Roost1513
1 hour ago
@Roost1513 You are welcome! Bye
– gimusi
1 hour ago
@Roost1513 A similar case to be considered is $lim_xto 0 frac x x$. In that case we have an indeterminate form $0/0$ at $x=0$ but for $xneq 0$ the function $f(x)=x/x=1$. Can you conclude what the limit is?
– gimusi
1 hour ago
 |Â
show 6 more comments
up vote
4
down vote
accepted
Note that for any $x$ we have $xcdot 0=0$ and therefore
$$lim_xtoinfty (xcdot 0) =lim_xtoinfty 0=0$$
answered 2 hours ago
gimusi
78.4k73889
Would this change if I didn't use parenthesis?
– Roost1513
1 hour ago
@Roost1513 I've used the parenthesis just to indicate that we are taking the limit for $(xcdot 0)$ which is the identically zero function $f(x)=0$. For that reason the limit is equal to $0$.
– gimusi
1 hour ago
Thank you for the clarification, I just wanted to be absolutely certain.
– Roost1513
1 hour ago
@Roost1513 You are welcome! Bye
– gimusi
1 hour ago
@Roost1513 A similar case to be considered is $lim_xto 0 frac x x$. In that case we have an indeterminate form $0/0$ at $x=0$ but for $xneq 0$ the function $f(x)=x/x=1$. Can you conclude what the limit is?
– gimusi
1 hour ago
 |Â
show 6 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Note that for any $x$ we have $xcdot 0=0$ and therefore
$$lim_xtoinfty (xcdot 0) =lim_xtoinfty 0=0$$
answered 2 hours ago
gimusi
78.4k73889
Note that for any $x$ we have $xcdot 0=0$ and therefore
$$lim_xtoinfty (xcdot 0) =lim_xtoinfty 0=0$$
answered 2 hours ago
gimusi
78.4k73889
answered 2 hours ago
gimusi
78.4k73889
answered 2 hours ago
gimusi
78.4k73889
answered 2 hours ago
gimusi
78.4k73889
78.4k73889
Would this change if I didn't use parenthesis?
– Roost1513
1 hour ago
@Roost1513 I've used the parenthesis just to indicate that we are taking the limit for $(xcdot 0)$ which is the identically zero function $f(x)=0$. For that reason the limit is equal to $0$.
– gimusi
1 hour ago
Thank you for the clarification, I just wanted to be absolutely certain.
– Roost1513
1 hour ago
@Roost1513 You are welcome! Bye
– gimusi
1 hour ago
@Roost1513 A similar case to be considered is $lim_xto 0 frac x x$. In that case we have an indeterminate form $0/0$ at $x=0$ but for $xneq 0$ the function $f(x)=x/x=1$. Can you conclude what the limit is?
– gimusi
1 hour ago
 |Â
show 6 more comments
Would this change if I didn't use parenthesis?
– Roost1513
1 hour ago
@Roost1513 I've used the parenthesis just to indicate that we are taking the limit for $(xcdot 0)$ which is the identically zero function $f(x)=0$. For that reason the limit is equal to $0$.
– gimusi
1 hour ago
Thank you for the clarification, I just wanted to be absolutely certain.
– Roost1513
1 hour ago
@Roost1513 You are welcome! Bye
– gimusi
1 hour ago
@Roost1513 A similar case to be considered is $lim_xto 0 frac x x$. In that case we have an indeterminate form $0/0$ at $x=0$ but for $xneq 0$ the function $f(x)=x/x=1$. Can you conclude what the limit is?
– gimusi
1 hour ago
Would this change if I didn't use parenthesis?
– Roost1513
1 hour ago
Would this change if I didn't use parenthesis?
– Roost1513
1 hour ago
@Roost1513 I've used the parenthesis just to indicate that we are taking the limit for $(xcdot 0)$ which is the identically zero function $f(x)=0$. For that reason the limit is equal to $0$.
– gimusi
1 hour ago
@Roost1513 I've used the parenthesis just to indicate that we are taking the limit for $(xcdot 0)$ which is the identically zero function $f(x)=0$. For that reason the limit is equal to $0$.
– gimusi
1 hour ago
Thank you for the clarification, I just wanted to be absolutely certain.
– Roost1513
1 hour ago
Thank you for the clarification, I just wanted to be absolutely certain.
– Roost1513
1 hour ago
@Roost1513 You are welcome! Bye
– gimusi
1 hour ago
@Roost1513 You are welcome! Bye
– gimusi
1 hour ago
@Roost1513 A similar case to be considered is $lim_xto 0 frac x x$. In that case we have an indeterminate form $0/0$ at $x=0$ but for $xneq 0$ the function $f(x)=x/x=1$. Can you conclude what the limit is?
– gimusi
1 hour ago
@Roost1513 A similar case to be considered is $lim_xto 0 frac x x$. In that case we have an indeterminate form $0/0$ at $x=0$ but for $xneq 0$ the function $f(x)=x/x=1$. Can you conclude what the limit is?
– gimusi
1 hour ago
 |Â
show 6 more comments
up vote
2
down vote
For every $xin mathbf R$ it holds $0 cdot x = 0$. Using this definition, we let $f(x):= 0 cdot x = 0$ for every $xin mathbf R$. Hence, we have
$$ lim_xto +infty xcdot 0 = lim_xto +infty f(x) = lim_xto +infty 0 =0$$
from the definition of the limit! There is no need to think about something like $0 cdot (+infty)$.
answered 2 hours ago
Niklas
2,093719
add a comment |Â
up vote
2
down vote
For every $xin mathbf R$ it holds $0 cdot x = 0$. Using this definition, we let $f(x):= 0 cdot x = 0$ for every $xin mathbf R$. Hence, we have
$$ lim_xto +infty xcdot 0 = lim_xto +infty f(x) = lim_xto +infty 0 =0$$
from the definition of the limit! There is no need to think about something like $0 cdot (+infty)$.
answered 2 hours ago
Niklas
2,093719
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For every $xin mathbf R$ it holds $0 cdot x = 0$. Using this definition, we let $f(x):= 0 cdot x = 0$ for every $xin mathbf R$. Hence, we have
$$ lim_xto +infty xcdot 0 = lim_xto +infty f(x) = lim_xto +infty 0 =0$$
from the definition of the limit! There is no need to think about something like $0 cdot (+infty)$.
answered 2 hours ago
Niklas
2,093719
For every $xin mathbf R$ it holds $0 cdot x = 0$. Using this definition, we let $f(x):= 0 cdot x = 0$ for every $xin mathbf R$. Hence, we have
$$ lim_xto +infty xcdot 0 = lim_xto +infty f(x) = lim_xto +infty 0 =0$$
from the definition of the limit! There is no need to think about something like $0 cdot (+infty)$.
answered 2 hours ago
Niklas
2,093719
answered 2 hours ago
Niklas
2,093719
answered 2 hours ago
Niklas
2,093719
answered 2 hours ago
Niklas
2,093719
2,093719
add a comment |Â
add a comment |Â
up vote
0
down vote
The limit is zero.
The reason is as follows. The function you are considering is $f(x) = x times 0$. But this means that $f(x)=0$ for all real $x$. The limit of this function as x tends to infinity is 0, even though as you point out $0timesinfty$ is undefined (but we do not need to calculate that here).
Formally, to show that this limit is zero, we need to show that for all $epsilon>0$ there exists a real $N$ so that $|f(x)-0|<epsilon$ for all $xge N$. But this is trivially true for any real $N$ and any $epsilon>0$.
answered 2 hours ago
Stefanie
11
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Stefanie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
The limit is zero.
The reason is as follows. The function you are considering is $f(x) = x times 0$. But this means that $f(x)=0$ for all real $x$. The limit of this function as x tends to infinity is 0, even though as you point out $0timesinfty$ is undefined (but we do not need to calculate that here).
Formally, to show that this limit is zero, we need to show that for all $epsilon>0$ there exists a real $N$ so that $|f(x)-0|<epsilon$ for all $xge N$. But this is trivially true for any real $N$ and any $epsilon>0$.
answered 2 hours ago
Stefanie
11
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Stefanie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
up vote
0
down vote
The limit is zero.
The reason is as follows. The function you are considering is $f(x) = x times 0$. But this means that $f(x)=0$ for all real $x$. The limit of this function as x tends to infinity is 0, even though as you point out $0timesinfty$ is undefined (but we do not need to calculate that here).
Formally, to show that this limit is zero, we need to show that for all $epsilon>0$ there exists a real $N$ so that $|f(x)-0|<epsilon$ for all $xge N$. But this is trivially true for any real $N$ and any $epsilon>0$.
answered 2 hours ago
Stefanie
11
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Stefanie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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The limit is zero.
The reason is as follows. The function you are considering is $f(x) = x times 0$. But this means that $f(x)=0$ for all real $x$. The limit of this function as x tends to infinity is 0, even though as you point out $0timesinfty$ is undefined (but we do not need to calculate that here).
Formally, to show that this limit is zero, we need to show that for all $epsilon>0$ there exists a real $N$ so that $|f(x)-0|<epsilon$ for all $xge N$. But this is trivially true for any real $N$ and any $epsilon>0$.
answered 2 hours ago
Stefanie
11
New contributor
Stefanie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Stefanie
11
New contributor
Stefanie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 2 hours ago
Stefanie
11
answered 2 hours ago
Stefanie
11
11
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Stefanie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Stefanie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Stefanie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |Â
up vote
0
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Let $f(x)=0times x$.
Then, the following statement is true:
For every $epsilon > 0$, there exists some $Min mathbb R$ such that, for all $x>M$, $|f(x) - 0| < epsilon$.
Therefore, by the definition of a limit, we can conclude that
$$lim_xtoinfty f(x) = 0$$
If you want the proof of the statement in yellow above:
Let $epsilon > 0$ be arbitrary. Then, let $M=1$. Let $x>M$ be arbitrary. Then, $|f(x)-0| = |0times x - 0| = |0-0|=0<epsilon$.
answered 2 hours ago
5xum
85.8k388154
add a comment |Â
up vote
0
down vote
Let $f(x)=0times x$.
Then, the following statement is true:
For every $epsilon > 0$, there exists some $Min mathbb R$ such that, for all $x>M$, $|f(x) - 0| < epsilon$.
Therefore, by the definition of a limit, we can conclude that
$$lim_xtoinfty f(x) = 0$$
If you want the proof of the statement in yellow above:
Let $epsilon > 0$ be arbitrary. Then, let $M=1$. Let $x>M$ be arbitrary. Then, $|f(x)-0| = |0times x - 0| = |0-0|=0<epsilon$.
answered 2 hours ago
5xum
85.8k388154
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $f(x)=0times x$.
Then, the following statement is true:
For every $epsilon > 0$, there exists some $Min mathbb R$ such that, for all $x>M$, $|f(x) - 0| < epsilon$.
Therefore, by the definition of a limit, we can conclude that
$$lim_xtoinfty f(x) = 0$$
If you want the proof of the statement in yellow above:
Let $epsilon > 0$ be arbitrary. Then, let $M=1$. Let $x>M$ be arbitrary. Then, $|f(x)-0| = |0times x - 0| = |0-0|=0<epsilon$.
answered 2 hours ago
5xum
85.8k388154
Let $f(x)=0times x$.
Then, the following statement is true:
For every $epsilon > 0$, there exists some $Min mathbb R$ such that, for all $x>M$, $|f(x) - 0| < epsilon$.
Therefore, by the definition of a limit, we can conclude that
$$lim_xtoinfty f(x) = 0$$
If you want the proof of the statement in yellow above:
Let $epsilon > 0$ be arbitrary. Then, let $M=1$. Let $x>M$ be arbitrary. Then, $|f(x)-0| = |0times x - 0| = |0-0|=0<epsilon$.
answered 2 hours ago
5xum
85.8k388154
answered 2 hours ago
5xum
85.8k388154
answered 2 hours ago
5xum
85.8k388154
answered 2 hours ago
5xum
85.8k388154
85.8k388154
add a comment |Â
add a comment |Â
up vote
0
down vote
As others have said, $lim_xto infty 0 times x = lim_xtoinfty 0 = 0$. I'm going to expand a bit more on "$0 times infty$ is undefined".
We can't do operations with $infty$ directly, as you know. But we can do operations with "functions with limit $infty$", and if they behave well enough then that might give us reasonable definitions of things like "$0 times infty$".
However, if we replace $infty$ by "functions with limit $infty$" then we ought to do the same with $0$, i.e. replace $0$ by "functions with limit $0$". This is a reasonable thing to do, because it works for real numbers:
For any functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x) = a$ and $lim_x to infty g(x) = b$, where $a$ and $b$ are real numbers (i.e. finite), it's the case that $lim_x to infty (f(x)g(x)) = ab$.
In fact, it's better than that; if $a > 0$ is real then it's reasonable to say "$a times infty = infty$":
For any functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x) = a$, where $a$ is a real number and $a > 0$, and such that $lim_x to infty g(x) = infty$, then $lim_x to infty(f(x)g(x)) = infty$ also.
However, if $f(x)$ and $g(x)$ are such that $lim_x to infty f(x) = 0$ and $lim _x to infty g(x) = infty$, then we don't know anything about $lim_x to infty f(x)g(x)$. If $g(x) = x$, then taking $f(x) = 0$, $f(x) = a/x$ (where $a > 0$), and $f(x) = 1/sqrt x$, gives $lim_x to infty (f(x)g(x))$ to be, respectively, $0$, $a$ and $infty$. This is why we say "$0 times infty$" is undefined.
answered 1 hour ago
Christopher
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As others have said, $lim_xto infty 0 times x = lim_xtoinfty 0 = 0$. I'm going to expand a bit more on "$0 times infty$ is undefined".
We can't do operations with $infty$ directly, as you know. But we can do operations with "functions with limit $infty$", and if they behave well enough then that might give us reasonable definitions of things like "$0 times infty$".
However, if we replace $infty$ by "functions with limit $infty$" then we ought to do the same with $0$, i.e. replace $0$ by "functions with limit $0$". This is a reasonable thing to do, because it works for real numbers:
For any functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x) = a$ and $lim_x to infty g(x) = b$, where $a$ and $b$ are real numbers (i.e. finite), it's the case that $lim_x to infty (f(x)g(x)) = ab$.
In fact, it's better than that; if $a > 0$ is real then it's reasonable to say "$a times infty = infty$":
For any functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x) = a$, where $a$ is a real number and $a > 0$, and such that $lim_x to infty g(x) = infty$, then $lim_x to infty(f(x)g(x)) = infty$ also.
However, if $f(x)$ and $g(x)$ are such that $lim_x to infty f(x) = 0$ and $lim _x to infty g(x) = infty$, then we don't know anything about $lim_x to infty f(x)g(x)$. If $g(x) = x$, then taking $f(x) = 0$, $f(x) = a/x$ (where $a > 0$), and $f(x) = 1/sqrt x$, gives $lim_x to infty (f(x)g(x))$ to be, respectively, $0$, $a$ and $infty$. This is why we say "$0 times infty$" is undefined.
answered 1 hour ago
Christopher
5,59911527
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As others have said, $lim_xto infty 0 times x = lim_xtoinfty 0 = 0$. I'm going to expand a bit more on "$0 times infty$ is undefined".
We can't do operations with $infty$ directly, as you know. But we can do operations with "functions with limit $infty$", and if they behave well enough then that might give us reasonable definitions of things like "$0 times infty$".
However, if we replace $infty$ by "functions with limit $infty$" then we ought to do the same with $0$, i.e. replace $0$ by "functions with limit $0$". This is a reasonable thing to do, because it works for real numbers:
For any functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x) = a$ and $lim_x to infty g(x) = b$, where $a$ and $b$ are real numbers (i.e. finite), it's the case that $lim_x to infty (f(x)g(x)) = ab$.
In fact, it's better than that; if $a > 0$ is real then it's reasonable to say "$a times infty = infty$":
For any functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x) = a$, where $a$ is a real number and $a > 0$, and such that $lim_x to infty g(x) = infty$, then $lim_x to infty(f(x)g(x)) = infty$ also.
However, if $f(x)$ and $g(x)$ are such that $lim_x to infty f(x) = 0$ and $lim _x to infty g(x) = infty$, then we don't know anything about $lim_x to infty f(x)g(x)$. If $g(x) = x$, then taking $f(x) = 0$, $f(x) = a/x$ (where $a > 0$), and $f(x) = 1/sqrt x$, gives $lim_x to infty (f(x)g(x))$ to be, respectively, $0$, $a$ and $infty$. This is why we say "$0 times infty$" is undefined.
answered 1 hour ago
Christopher
5,59911527
As others have said, $lim_xto infty 0 times x = lim_xtoinfty 0 = 0$. I'm going to expand a bit more on "$0 times infty$ is undefined".
We can't do operations with $infty$ directly, as you know. But we can do operations with "functions with limit $infty$", and if they behave well enough then that might give us reasonable definitions of things like "$0 times infty$".
However, if we replace $infty$ by "functions with limit $infty$" then we ought to do the same with $0$, i.e. replace $0$ by "functions with limit $0$". This is a reasonable thing to do, because it works for real numbers:
For any functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x) = a$ and $lim_x to infty g(x) = b$, where $a$ and $b$ are real numbers (i.e. finite), it's the case that $lim_x to infty (f(x)g(x)) = ab$.
In fact, it's better than that; if $a > 0$ is real then it's reasonable to say "$a times infty = infty$":
For any functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x) = a$, where $a$ is a real number and $a > 0$, and such that $lim_x to infty g(x) = infty$, then $lim_x to infty(f(x)g(x)) = infty$ also.
However, if $f(x)$ and $g(x)$ are such that $lim_x to infty f(x) = 0$ and $lim _x to infty g(x) = infty$, then we don't know anything about $lim_x to infty f(x)g(x)$. If $g(x) = x$, then taking $f(x) = 0$, $f(x) = a/x$ (where $a > 0$), and $f(x) = 1/sqrt x$, gives $lim_x to infty (f(x)g(x))$ to be, respectively, $0$, $a$ and $infty$. This is why we say "$0 times infty$" is undefined.
answered 1 hour ago
Christopher
5,59911527
answered 1 hour ago
Christopher
5,59911527
answered 1 hour ago
Christopher
5,59911527
answered 1 hour ago
Christopher
5,59911527
5,59911527
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add a comment |Â
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Answer to the first limit is 0. The answer is 0 because we apply the L-hospital rule and differentiate the function.
answered 2 hours ago
pinkhelium9
51
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1
L’Hopital rule?
– gimusi
2 hours ago
1
You don't need the L'Hospital rule, and in fact, its use just obfuscates the solution.
– 5xum
2 hours ago
add a comment |Â
up vote
-2
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Answer to the first limit is 0. The answer is 0 because we apply the L-hospital rule and differentiate the function.
answered 2 hours ago
pinkhelium9
51
New contributor
pinkhelium9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
L’Hopital rule?
– gimusi
2 hours ago
1
You don't need the L'Hospital rule, and in fact, its use just obfuscates the solution.
– 5xum
2 hours ago
add a comment |Â
up vote
-2
down vote
up vote
-2
down vote
Answer to the first limit is 0. The answer is 0 because we apply the L-hospital rule and differentiate the function.
answered 2 hours ago
pinkhelium9
51
New contributor
pinkhelium9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Answer to the first limit is 0. The answer is 0 because we apply the L-hospital rule and differentiate the function.
answered 2 hours ago
pinkhelium9
51
New contributor
pinkhelium9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
pinkhelium9
51
New contributor
pinkhelium9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
pinkhelium9
51
answered 2 hours ago
pinkhelium9
51
51
New contributor
pinkhelium9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
pinkhelium9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
pinkhelium9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
L’Hopital rule?
– gimusi
2 hours ago
1
You don't need the L'Hospital rule, and in fact, its use just obfuscates the solution.
– 5xum
2 hours ago
add a comment |Â
1
L’Hopital rule?
– gimusi
2 hours ago
1
You don't need the L'Hospital rule, and in fact, its use just obfuscates the solution.
– 5xum
2 hours ago
1
1
L’Hopital rule?
– gimusi
2 hours ago
L’Hopital rule?
– gimusi
2 hours ago
1
1
You don't need the L'Hospital rule, and in fact, its use just obfuscates the solution.
– 5xum
2 hours ago
You don't need the L'Hospital rule, and in fact, its use just obfuscates the solution.
– 5xum
2 hours ago
add a comment |Â
Roost1513 is a new contributor. Be nice, and check out our Code of Conduct.
Roost1513 is a new contributor. Be nice, and check out our Code of Conduct.
Roost1513 is a new contributor. Be nice, and check out our Code of Conduct.
Roost1513 is a new contributor. Be nice, and check out our Code of Conduct.
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1
Would need to evaluate the quantity in the bracket first, the way this is written. Then everything is identically zero.
– AnyAD
2 hours ago
Obviously can't apply limit laws to the product.
– AnyAD
2 hours ago