How to prove that $f(x)=lim_{atoinfty}e^{-ax}$ is not a function?
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$y=lim_{atoinfty}e^{-ax}$. How to prove that there does not exist a function $f$ such that $f(x)=y$?
calculus limits functions
asked Nov 15 at 23:32
High GPA
869419
add a comment |
up vote
-1
down vote
favorite
$y=lim_{atoinfty}e^{-ax}$. How to prove that there does not exist a function $f$ such that $f(x)=y$?
calculus limits functions
asked Nov 15 at 23:32
High GPA
869419
1
What's the value of the limit when x is negative?
– Mjiig
Nov 15 at 23:34
@Mjiig +inf. But how to show that it is not a function?
– High GPA
Nov 15 at 23:55
It is a function from the reals to the extended reals.
– Q the Platypus
Nov 16 at 0:55
Agreed, the required domain and codomain need specifying to give this a firm answer.
– The_Sympathizer
Nov 16 at 4:40
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
$y=lim_{atoinfty}e^{-ax}$. How to prove that there does not exist a function $f$ such that $f(x)=y$?
calculus limits functions
asked Nov 15 at 23:32
High GPA
869419
$y=lim_{atoinfty}e^{-ax}$. How to prove that there does not exist a function $f$ such that $f(x)=y$?
calculus limits functions
calculus limits functions
asked Nov 15 at 23:32
High GPA
869419
asked Nov 15 at 23:32
High GPA
869419
asked Nov 15 at 23:32
High GPA
869419
asked Nov 15 at 23:32
High GPA
869419
asked Nov 15 at 23:32
High GPA
869419
869419
1
What's the value of the limit when x is negative?
– Mjiig
Nov 15 at 23:34
@Mjiig +inf. But how to show that it is not a function?
– High GPA
Nov 15 at 23:55
It is a function from the reals to the extended reals.
– Q the Platypus
Nov 16 at 0:55
Agreed, the required domain and codomain need specifying to give this a firm answer.
– The_Sympathizer
Nov 16 at 4:40
add a comment |
1
What's the value of the limit when x is negative?
– Mjiig
Nov 15 at 23:34
@Mjiig +inf. But how to show that it is not a function?
– High GPA
Nov 15 at 23:55
It is a function from the reals to the extended reals.
– Q the Platypus
Nov 16 at 0:55
Agreed, the required domain and codomain need specifying to give this a firm answer.
– The_Sympathizer
Nov 16 at 4:40
1
1
What's the value of the limit when x is negative?
– Mjiig
Nov 15 at 23:34
What's the value of the limit when x is negative?
– Mjiig
Nov 15 at 23:34
@Mjiig +inf. But how to show that it is not a function?
– High GPA
Nov 15 at 23:55
@Mjiig +inf. But how to show that it is not a function?
– High GPA
Nov 15 at 23:55
It is a function from the reals to the extended reals.
– Q the Platypus
Nov 16 at 0:55
It is a function from the reals to the extended reals.
– Q the Platypus
Nov 16 at 0:55
Agreed, the required domain and codomain need specifying to give this a firm answer.
– The_Sympathizer
Nov 16 at 4:40
Agreed, the required domain and codomain need specifying to give this a firm answer.
– The_Sympathizer
Nov 16 at 4:40
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
Normally when we define a real-valued function, it takes values on $(-infty,+infty)$ unless we specify it to be extended real-valued, which means it can assume value $pminfty$.
answered Nov 16 at 0:25
Apocalypse
1065
I don't think the infinities are the issue here. Regardless of $x>0$ the limit always goes to $0$. The problem is when $x< 0$.
– YiFan
Nov 16 at 3:31
@YiFan I think the original problem is not specified. Clearly, the problem reduces to working out the limit for any given $x$. Then the limit just doesn't exist for $x<0$ (if we assume it not to be extended real-valued). If the original problem is "Show$ f(x)=lim_{atoinfty}e^{-ax} $is not an extended real-valued function", then it is clearly a wrong statement since we can define $f(x)=+infty$ for $x<0$, $f(x)=1$ for $x=0$ and $f(x)=0$ for $x>0$ .
– Apocalypse
Nov 16 at 3:35
add a comment |
up vote
0
down vote
Contrary to what you said, there are many ways this can be a function. The easiest way is to define it as a function from $mathbb R_{geq0}$ to $mathbb R$, which works with $f(x)=0$ for any $x>0$, and $f(0)=1$. You can also say that it is a function from $mathbb R$ to $overline{mathbb R}$, the extended real number system containing both infinities. Meanwhile, it is not a function from $mathbb Rtomathbb R$ because $f(-1)$, for example, is undefined. The limit goes to $infty$, but $infty$ is not a real number.
answered Nov 16 at 3:39
YiFan
1,5711312
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Normally when we define a real-valued function, it takes values on $(-infty,+infty)$ unless we specify it to be extended real-valued, which means it can assume value $pminfty$.
answered Nov 16 at 0:25
Apocalypse
1065
I don't think the infinities are the issue here. Regardless of $x>0$ the limit always goes to $0$. The problem is when $x< 0$.
– YiFan
Nov 16 at 3:31
@YiFan I think the original problem is not specified. Clearly, the problem reduces to working out the limit for any given $x$. Then the limit just doesn't exist for $x<0$ (if we assume it not to be extended real-valued). If the original problem is "Show$ f(x)=lim_{atoinfty}e^{-ax} $is not an extended real-valued function", then it is clearly a wrong statement since we can define $f(x)=+infty$ for $x<0$, $f(x)=1$ for $x=0$ and $f(x)=0$ for $x>0$ .
– Apocalypse
Nov 16 at 3:35
add a comment |
up vote
0
down vote
Normally when we define a real-valued function, it takes values on $(-infty,+infty)$ unless we specify it to be extended real-valued, which means it can assume value $pminfty$.
answered Nov 16 at 0:25
Apocalypse
1065
I don't think the infinities are the issue here. Regardless of $x>0$ the limit always goes to $0$. The problem is when $x< 0$.
– YiFan
Nov 16 at 3:31
@YiFan I think the original problem is not specified. Clearly, the problem reduces to working out the limit for any given $x$. Then the limit just doesn't exist for $x<0$ (if we assume it not to be extended real-valued). If the original problem is "Show$ f(x)=lim_{atoinfty}e^{-ax} $is not an extended real-valued function", then it is clearly a wrong statement since we can define $f(x)=+infty$ for $x<0$, $f(x)=1$ for $x=0$ and $f(x)=0$ for $x>0$ .
– Apocalypse
Nov 16 at 3:35
add a comment |
up vote
0
down vote
up vote
0
down vote
Normally when we define a real-valued function, it takes values on $(-infty,+infty)$ unless we specify it to be extended real-valued, which means it can assume value $pminfty$.
answered Nov 16 at 0:25
Apocalypse
1065
Normally when we define a real-valued function, it takes values on $(-infty,+infty)$ unless we specify it to be extended real-valued, which means it can assume value $pminfty$.
answered Nov 16 at 0:25
Apocalypse
1065
edited Nov 16 at 0:34
answered Nov 16 at 0:25
Apocalypse
1065
answered Nov 16 at 0:25
Apocalypse
1065
answered Nov 16 at 0:25
Apocalypse
1065
1065
I don't think the infinities are the issue here. Regardless of $x>0$ the limit always goes to $0$. The problem is when $x< 0$.
– YiFan
Nov 16 at 3:31
@YiFan I think the original problem is not specified. Clearly, the problem reduces to working out the limit for any given $x$. Then the limit just doesn't exist for $x<0$ (if we assume it not to be extended real-valued). If the original problem is "Show$ f(x)=lim_{atoinfty}e^{-ax} $is not an extended real-valued function", then it is clearly a wrong statement since we can define $f(x)=+infty$ for $x<0$, $f(x)=1$ for $x=0$ and $f(x)=0$ for $x>0$ .
– Apocalypse
Nov 16 at 3:35
add a comment |
I don't think the infinities are the issue here. Regardless of $x>0$ the limit always goes to $0$. The problem is when $x< 0$.
– YiFan
Nov 16 at 3:31
@YiFan I think the original problem is not specified. Clearly, the problem reduces to working out the limit for any given $x$. Then the limit just doesn't exist for $x<0$ (if we assume it not to be extended real-valued). If the original problem is "Show$ f(x)=lim_{atoinfty}e^{-ax} $is not an extended real-valued function", then it is clearly a wrong statement since we can define $f(x)=+infty$ for $x<0$, $f(x)=1$ for $x=0$ and $f(x)=0$ for $x>0$ .
– Apocalypse
Nov 16 at 3:35
I don't think the infinities are the issue here. Regardless of $x>0$ the limit always goes to $0$. The problem is when $x< 0$.
– YiFan
Nov 16 at 3:31
I don't think the infinities are the issue here. Regardless of $x>0$ the limit always goes to $0$. The problem is when $x< 0$.
– YiFan
Nov 16 at 3:31
@YiFan I think the original problem is not specified. Clearly, the problem reduces to working out the limit for any given $x$. Then the limit just doesn't exist for $x<0$ (if we assume it not to be extended real-valued). If the original problem is "Show$ f(x)=lim_{atoinfty}e^{-ax} $is not an extended real-valued function", then it is clearly a wrong statement since we can define $f(x)=+infty$ for $x<0$, $f(x)=1$ for $x=0$ and $f(x)=0$ for $x>0$ .
– Apocalypse
Nov 16 at 3:35
@YiFan I think the original problem is not specified. Clearly, the problem reduces to working out the limit for any given $x$. Then the limit just doesn't exist for $x<0$ (if we assume it not to be extended real-valued). If the original problem is "Show$ f(x)=lim_{atoinfty}e^{-ax} $is not an extended real-valued function", then it is clearly a wrong statement since we can define $f(x)=+infty$ for $x<0$, $f(x)=1$ for $x=0$ and $f(x)=0$ for $x>0$ .
– Apocalypse
Nov 16 at 3:35
add a comment |
up vote
0
down vote
Contrary to what you said, there are many ways this can be a function. The easiest way is to define it as a function from $mathbb R_{geq0}$ to $mathbb R$, which works with $f(x)=0$ for any $x>0$, and $f(0)=1$. You can also say that it is a function from $mathbb R$ to $overline{mathbb R}$, the extended real number system containing both infinities. Meanwhile, it is not a function from $mathbb Rtomathbb R$ because $f(-1)$, for example, is undefined. The limit goes to $infty$, but $infty$ is not a real number.
answered Nov 16 at 3:39
YiFan
1,5711312
add a comment |
up vote
0
down vote
Contrary to what you said, there are many ways this can be a function. The easiest way is to define it as a function from $mathbb R_{geq0}$ to $mathbb R$, which works with $f(x)=0$ for any $x>0$, and $f(0)=1$. You can also say that it is a function from $mathbb R$ to $overline{mathbb R}$, the extended real number system containing both infinities. Meanwhile, it is not a function from $mathbb Rtomathbb R$ because $f(-1)$, for example, is undefined. The limit goes to $infty$, but $infty$ is not a real number.
answered Nov 16 at 3:39
YiFan
1,5711312
add a comment |
up vote
0
down vote
up vote
0
down vote
Contrary to what you said, there are many ways this can be a function. The easiest way is to define it as a function from $mathbb R_{geq0}$ to $mathbb R$, which works with $f(x)=0$ for any $x>0$, and $f(0)=1$. You can also say that it is a function from $mathbb R$ to $overline{mathbb R}$, the extended real number system containing both infinities. Meanwhile, it is not a function from $mathbb Rtomathbb R$ because $f(-1)$, for example, is undefined. The limit goes to $infty$, but $infty$ is not a real number.
answered Nov 16 at 3:39
YiFan
1,5711312
Contrary to what you said, there are many ways this can be a function. The easiest way is to define it as a function from $mathbb R_{geq0}$ to $mathbb R$, which works with $f(x)=0$ for any $x>0$, and $f(0)=1$. You can also say that it is a function from $mathbb R$ to $overline{mathbb R}$, the extended real number system containing both infinities. Meanwhile, it is not a function from $mathbb Rtomathbb R$ because $f(-1)$, for example, is undefined. The limit goes to $infty$, but $infty$ is not a real number.
answered Nov 16 at 3:39
YiFan
1,5711312
answered Nov 16 at 3:39
YiFan
1,5711312
answered Nov 16 at 3:39
YiFan
1,5711312
answered Nov 16 at 3:39
YiFan
1,5711312
1,5711312
add a comment |
add a comment |
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What's the value of the limit when x is negative?
– Mjiig
Nov 15 at 23:34
@Mjiig +inf. But how to show that it is not a function?
– High GPA
Nov 15 at 23:55
It is a function from the reals to the extended reals.
– Q the Platypus
Nov 16 at 0:55
Agreed, the required domain and codomain need specifying to give this a firm answer.
– The_Sympathizer
Nov 16 at 4:40