Mixing
Show $f(x)=frac{1}{|x|}$ is discontinuous at x=0
$begingroup$
I am struggling to explicitly show that the function is discontinuous at $x=0$ since the usual technique of finding two sequences $x_n$,$y_n$ where $x_nrightarrow0$, $y_nrightarrow0$ but $f(x_n)$ and $f(y_n)$ tend to different values, doesnt seem to work as all seuqnces tend to $+infty$.
My only other thought would be that since $f(0)$ doesn't exist, then from this $f(x)$ cannot be continuous. But I'm unsure if this would suffice as an answer.
real-analysis
edited May 21 '18 at 11:16
Siminore
30.4k33368
asked May 21 '18 at 11:15
DLBDLB
548
$endgroup$
add a comment |
$begingroup$
I am struggling to explicitly show that the function is discontinuous at $x=0$ since the usual technique of finding two sequences $x_n$,$y_n$ where $x_nrightarrow0$, $y_nrightarrow0$ but $f(x_n)$ and $f(y_n)$ tend to different values, doesnt seem to work as all seuqnces tend to $+infty$.
My only other thought would be that since $f(0)$ doesn't exist, then from this $f(x)$ cannot be continuous. But I'm unsure if this would suffice as an answer.
real-analysis
edited May 21 '18 at 11:16
Siminore
30.4k33368
asked May 21 '18 at 11:15
DLBDLB
548
$endgroup$
$begingroup$
The function $frac{1}{|x|}$ is not defined at $x=0$, and so you cannot ask whether it is continuous there. It is continuous on $mathbb{R}setminus {0}$.
$endgroup$
– Daniel Beale
May 21 '18 at 11:33
$begingroup$
Possible duplicate of Can we talk about the continuity/discontinuity of a function at a point which is not in its domain?
$endgroup$
– Michael Hoppe
May 21 '18 at 12:36
$begingroup$
Roughly speaking, the following ensemble of symbols and words: ‘$1/|x| colonmathbb R^*tomathbb R$ is not continuous in $0$.’ is no statement, i.e., neither true nor false.
$endgroup$
– Michael Hoppe
May 21 '18 at 12:40
add a comment |
$begingroup$
I am struggling to explicitly show that the function is discontinuous at $x=0$ since the usual technique of finding two sequences $x_n$,$y_n$ where $x_nrightarrow0$, $y_nrightarrow0$ but $f(x_n)$ and $f(y_n)$ tend to different values, doesnt seem to work as all seuqnces tend to $+infty$.
My only other thought would be that since $f(0)$ doesn't exist, then from this $f(x)$ cannot be continuous. But I'm unsure if this would suffice as an answer.
real-analysis
edited May 21 '18 at 11:16
Siminore
30.4k33368
asked May 21 '18 at 11:15
DLBDLB
548
$endgroup$
I am struggling to explicitly show that the function is discontinuous at $x=0$ since the usual technique of finding two sequences $x_n$,$y_n$ where $x_nrightarrow0$, $y_nrightarrow0$ but $f(x_n)$ and $f(y_n)$ tend to different values, doesnt seem to work as all seuqnces tend to $+infty$.
My only other thought would be that since $f(0)$ doesn't exist, then from this $f(x)$ cannot be continuous. But I'm unsure if this would suffice as an answer.
real-analysis
real-analysis
edited May 21 '18 at 11:16
Siminore
30.4k33368
asked May 21 '18 at 11:15
DLBDLB
548
edited May 21 '18 at 11:16
Siminore
30.4k33368
asked May 21 '18 at 11:15
DLBDLB
548
edited May 21 '18 at 11:16
Siminore
30.4k33368
edited May 21 '18 at 11:16
Siminore
30.4k33368
edited May 21 '18 at 11:16
Siminore
30.4k33368
30.4k33368
asked May 21 '18 at 11:15
DLBDLB
548
asked May 21 '18 at 11:15
DLBDLB
548
asked May 21 '18 at 11:15
DLBDLB
548
548
$begingroup$
The function $frac{1}{|x|}$ is not defined at $x=0$, and so you cannot ask whether it is continuous there. It is continuous on $mathbb{R}setminus {0}$.
$endgroup$
– Daniel Beale
May 21 '18 at 11:33
$begingroup$
Possible duplicate of Can we talk about the continuity/discontinuity of a function at a point which is not in its domain?
$endgroup$
– Michael Hoppe
May 21 '18 at 12:36
$begingroup$
Roughly speaking, the following ensemble of symbols and words: ‘$1/|x| colonmathbb R^*tomathbb R$ is not continuous in $0$.’ is no statement, i.e., neither true nor false.
$endgroup$
– Michael Hoppe
May 21 '18 at 12:40
add a comment |
$begingroup$
The function $frac{1}{|x|}$ is not defined at $x=0$, and so you cannot ask whether it is continuous there. It is continuous on $mathbb{R}setminus {0}$.
$endgroup$
– Daniel Beale
May 21 '18 at 11:33
$begingroup$
Possible duplicate of Can we talk about the continuity/discontinuity of a function at a point which is not in its domain?
$endgroup$
– Michael Hoppe
May 21 '18 at 12:36
$begingroup$
Roughly speaking, the following ensemble of symbols and words: ‘$1/|x| colonmathbb R^*tomathbb R$ is not continuous in $0$.’ is no statement, i.e., neither true nor false.
$endgroup$
– Michael Hoppe
May 21 '18 at 12:40
$begingroup$
The function $frac{1}{|x|}$ is not defined at $x=0$, and so you cannot ask whether it is continuous there. It is continuous on $mathbb{R}setminus {0}$.
$endgroup$
– Daniel Beale
May 21 '18 at 11:33
$begingroup$
The function $frac{1}{|x|}$ is not defined at $x=0$, and so you cannot ask whether it is continuous there. It is continuous on $mathbb{R}setminus {0}$.
$endgroup$
– Daniel Beale
May 21 '18 at 11:33
$begingroup$
Possible duplicate of Can we talk about the continuity/discontinuity of a function at a point which is not in its domain?
$endgroup$
– Michael Hoppe
May 21 '18 at 12:36
$begingroup$
Possible duplicate of Can we talk about the continuity/discontinuity of a function at a point which is not in its domain?
$endgroup$
– Michael Hoppe
May 21 '18 at 12:36
$begingroup$
Roughly speaking, the following ensemble of symbols and words: ‘$1/|x| colonmathbb R^*tomathbb R$ is not continuous in $0$.’ is no statement, i.e., neither true nor false.
$endgroup$
– Michael Hoppe
May 21 '18 at 12:40
$begingroup$
Roughly speaking, the following ensemble of symbols and words: ‘$1/|x| colonmathbb R^*tomathbb R$ is not continuous in $0$.’ is no statement, i.e., neither true nor false.
$endgroup$
– Michael Hoppe
May 21 '18 at 12:40
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If my students dare to say that this function is discontinuous at $0$, they will miserably fail. But we all know that the definition of discontinuity is somehow vague.
- Many mathematicians clearly state that a function can be discontinuous only at points of the domain. For these people, your work is useless, since your function is continuous throughout the domain of definition.
- Other mathematicians say that a function is always discontinuous at points that do not belong to the domain of definition, but they are accumulation points for the domain. Indeed, since $f(x_0)=lim_{x to x_0}f(x)$ is the characterizing property of continuous functions, it is violated as soon as $f(x_0)$ does not exist. Even in this case your work is useless, since your function is automatically continuous.
I won't write here a long discussion about the "best" definition of discontinuity, since it has been done several times in the past.
answered May 21 '18 at 11:21
SiminoreSiminore
30.4k33368
$endgroup$
add a comment |
$begingroup$
This is what I was taught in high school: For continuity at $x=a$ (this being a limit point), the function $f(x)$ must satisfy the condition $$lim_{xto a^+}f(x)=lim_{xto a^-}f(x)=f(a)$$ and the limit must exist.
In this case, $f(x)=dfrac 1{|x|}$ is not even defined at $x=0$, and hence is not continuous. Not only that, but the limit $lim_{xto a}f(x)=infty$ and hence does not exist.
answered May 21 '18 at 11:22
IshanIshan
2,0221927
$endgroup$
add a comment |
$begingroup$
Let $displaystyle x_n=frac{1}{n}$ then $displaystyle x_n underset{n rightarrow +infty}{rightarrow}0$ and
$$
fleft(x_nright) underset{n rightarrow +infty}{rightarrow}+infty
$$
Hence $f$ is not continuous at $x=0$
answered May 21 '18 at 11:19
AtmosAtmos
4,805420
$endgroup$
1
$begingroup$
But $f(x)=frac{1}{|x|}$ is not definied for $x=0$ so your question is senseless
$endgroup$
– Dr. Sonnhard Graubner
May 21 '18 at 11:22
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If my students dare to say that this function is discontinuous at $0$, they will miserably fail. But we all know that the definition of discontinuity is somehow vague.
- Many mathematicians clearly state that a function can be discontinuous only at points of the domain. For these people, your work is useless, since your function is continuous throughout the domain of definition.
- Other mathematicians say that a function is always discontinuous at points that do not belong to the domain of definition, but they are accumulation points for the domain. Indeed, since $f(x_0)=lim_{x to x_0}f(x)$ is the characterizing property of continuous functions, it is violated as soon as $f(x_0)$ does not exist. Even in this case your work is useless, since your function is automatically continuous.
I won't write here a long discussion about the "best" definition of discontinuity, since it has been done several times in the past.
answered May 21 '18 at 11:21
SiminoreSiminore
30.4k33368
$endgroup$
add a comment |
$begingroup$
If my students dare to say that this function is discontinuous at $0$, they will miserably fail. But we all know that the definition of discontinuity is somehow vague.
- Many mathematicians clearly state that a function can be discontinuous only at points of the domain. For these people, your work is useless, since your function is continuous throughout the domain of definition.
- Other mathematicians say that a function is always discontinuous at points that do not belong to the domain of definition, but they are accumulation points for the domain. Indeed, since $f(x_0)=lim_{x to x_0}f(x)$ is the characterizing property of continuous functions, it is violated as soon as $f(x_0)$ does not exist. Even in this case your work is useless, since your function is automatically continuous.
I won't write here a long discussion about the "best" definition of discontinuity, since it has been done several times in the past.
answered May 21 '18 at 11:21
SiminoreSiminore
30.4k33368
$endgroup$
add a comment |
$begingroup$
If my students dare to say that this function is discontinuous at $0$, they will miserably fail. But we all know that the definition of discontinuity is somehow vague.
- Many mathematicians clearly state that a function can be discontinuous only at points of the domain. For these people, your work is useless, since your function is continuous throughout the domain of definition.
- Other mathematicians say that a function is always discontinuous at points that do not belong to the domain of definition, but they are accumulation points for the domain. Indeed, since $f(x_0)=lim_{x to x_0}f(x)$ is the characterizing property of continuous functions, it is violated as soon as $f(x_0)$ does not exist. Even in this case your work is useless, since your function is automatically continuous.
I won't write here a long discussion about the "best" definition of discontinuity, since it has been done several times in the past.
answered May 21 '18 at 11:21
SiminoreSiminore
30.4k33368
$endgroup$
If my students dare to say that this function is discontinuous at $0$, they will miserably fail. But we all know that the definition of discontinuity is somehow vague.
- Many mathematicians clearly state that a function can be discontinuous only at points of the domain. For these people, your work is useless, since your function is continuous throughout the domain of definition.
- Other mathematicians say that a function is always discontinuous at points that do not belong to the domain of definition, but they are accumulation points for the domain. Indeed, since $f(x_0)=lim_{x to x_0}f(x)$ is the characterizing property of continuous functions, it is violated as soon as $f(x_0)$ does not exist. Even in this case your work is useless, since your function is automatically continuous.
I won't write here a long discussion about the "best" definition of discontinuity, since it has been done several times in the past.
answered May 21 '18 at 11:21
SiminoreSiminore
30.4k33368
answered May 21 '18 at 11:21
SiminoreSiminore
30.4k33368
answered May 21 '18 at 11:21
SiminoreSiminore
30.4k33368
answered May 21 '18 at 11:21
SiminoreSiminore
30.4k33368
30.4k33368
add a comment |
add a comment |
$begingroup$
This is what I was taught in high school: For continuity at $x=a$ (this being a limit point), the function $f(x)$ must satisfy the condition $$lim_{xto a^+}f(x)=lim_{xto a^-}f(x)=f(a)$$ and the limit must exist.
In this case, $f(x)=dfrac 1{|x|}$ is not even defined at $x=0$, and hence is not continuous. Not only that, but the limit $lim_{xto a}f(x)=infty$ and hence does not exist.
answered May 21 '18 at 11:22
IshanIshan
2,0221927
$endgroup$
add a comment |
$begingroup$
This is what I was taught in high school: For continuity at $x=a$ (this being a limit point), the function $f(x)$ must satisfy the condition $$lim_{xto a^+}f(x)=lim_{xto a^-}f(x)=f(a)$$ and the limit must exist.
In this case, $f(x)=dfrac 1{|x|}$ is not even defined at $x=0$, and hence is not continuous. Not only that, but the limit $lim_{xto a}f(x)=infty$ and hence does not exist.
answered May 21 '18 at 11:22
IshanIshan
2,0221927
$endgroup$
add a comment |
$begingroup$
This is what I was taught in high school: For continuity at $x=a$ (this being a limit point), the function $f(x)$ must satisfy the condition $$lim_{xto a^+}f(x)=lim_{xto a^-}f(x)=f(a)$$ and the limit must exist.
In this case, $f(x)=dfrac 1{|x|}$ is not even defined at $x=0$, and hence is not continuous. Not only that, but the limit $lim_{xto a}f(x)=infty$ and hence does not exist.
answered May 21 '18 at 11:22
IshanIshan
2,0221927
$endgroup$
This is what I was taught in high school: For continuity at $x=a$ (this being a limit point), the function $f(x)$ must satisfy the condition $$lim_{xto a^+}f(x)=lim_{xto a^-}f(x)=f(a)$$ and the limit must exist.
In this case, $f(x)=dfrac 1{|x|}$ is not even defined at $x=0$, and hence is not continuous. Not only that, but the limit $lim_{xto a}f(x)=infty$ and hence does not exist.
answered May 21 '18 at 11:22
IshanIshan
2,0221927
edited Jan 11 at 7:08
answered May 21 '18 at 11:22
IshanIshan
2,0221927
answered May 21 '18 at 11:22
IshanIshan
2,0221927
answered May 21 '18 at 11:22
IshanIshan
2,0221927
2,0221927
add a comment |
add a comment |
$begingroup$
Let $displaystyle x_n=frac{1}{n}$ then $displaystyle x_n underset{n rightarrow +infty}{rightarrow}0$ and
$$
fleft(x_nright) underset{n rightarrow +infty}{rightarrow}+infty
$$
Hence $f$ is not continuous at $x=0$
answered May 21 '18 at 11:19
AtmosAtmos
4,805420
$endgroup$
1
$begingroup$
But $f(x)=frac{1}{|x|}$ is not definied for $x=0$ so your question is senseless
$endgroup$
– Dr. Sonnhard Graubner
May 21 '18 at 11:22
add a comment |
$begingroup$
Let $displaystyle x_n=frac{1}{n}$ then $displaystyle x_n underset{n rightarrow +infty}{rightarrow}0$ and
$$
fleft(x_nright) underset{n rightarrow +infty}{rightarrow}+infty
$$
Hence $f$ is not continuous at $x=0$
answered May 21 '18 at 11:19
AtmosAtmos
4,805420
$endgroup$
1
$begingroup$
But $f(x)=frac{1}{|x|}$ is not definied for $x=0$ so your question is senseless
$endgroup$
– Dr. Sonnhard Graubner
May 21 '18 at 11:22
add a comment |
$begingroup$
Let $displaystyle x_n=frac{1}{n}$ then $displaystyle x_n underset{n rightarrow +infty}{rightarrow}0$ and
$$
fleft(x_nright) underset{n rightarrow +infty}{rightarrow}+infty
$$
Hence $f$ is not continuous at $x=0$
answered May 21 '18 at 11:19
AtmosAtmos
4,805420
$endgroup$
Let $displaystyle x_n=frac{1}{n}$ then $displaystyle x_n underset{n rightarrow +infty}{rightarrow}0$ and
$$
fleft(x_nright) underset{n rightarrow +infty}{rightarrow}+infty
$$
Hence $f$ is not continuous at $x=0$
answered May 21 '18 at 11:19
AtmosAtmos
4,805420
answered May 21 '18 at 11:19
AtmosAtmos
4,805420
answered May 21 '18 at 11:19
AtmosAtmos
4,805420
answered May 21 '18 at 11:19
AtmosAtmos
4,805420
4,805420
1
$begingroup$
But $f(x)=frac{1}{|x|}$ is not definied for $x=0$ so your question is senseless
$endgroup$
– Dr. Sonnhard Graubner
May 21 '18 at 11:22
add a comment |
1
$begingroup$
But $f(x)=frac{1}{|x|}$ is not definied for $x=0$ so your question is senseless
$endgroup$
– Dr. Sonnhard Graubner
May 21 '18 at 11:22
1
1
$begingroup$
But $f(x)=frac{1}{|x|}$ is not definied for $x=0$ so your question is senseless
$endgroup$
– Dr. Sonnhard Graubner
May 21 '18 at 11:22
$begingroup$
But $f(x)=frac{1}{|x|}$ is not definied for $x=0$ so your question is senseless
$endgroup$
– Dr. Sonnhard Graubner
May 21 '18 at 11:22
add a comment |
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$begingroup$
The function $frac{1}{|x|}$ is not defined at $x=0$, and so you cannot ask whether it is continuous there. It is continuous on $mathbb{R}setminus {0}$.
$endgroup$
– Daniel Beale
May 21 '18 at 11:33
$begingroup$
Possible duplicate of Can we talk about the continuity/discontinuity of a function at a point which is not in its domain?
$endgroup$
– Michael Hoppe
May 21 '18 at 12:36
$begingroup$
Roughly speaking, the following ensemble of symbols and words: ‘$1/|x| colonmathbb R^*tomathbb R$ is not continuous in $0$.’ is no statement, i.e., neither true nor false.
$endgroup$
– Michael Hoppe
May 21 '18 at 12:40