Mixing
Is $sqrt x$ continuous at $0$? Because it is not defined to the left of $0$
$begingroup$
If a function has a limit from the right but not from the left, is it still continuous?
calculus
asked Sep 5 '16 at 3:28
Rodrigo StvRodrigo Stv
10317
$endgroup$
add a comment |
$begingroup$
If a function has a limit from the right but not from the left, is it still continuous?
calculus
asked Sep 5 '16 at 3:28
Rodrigo StvRodrigo Stv
10317
$endgroup$
$begingroup$
No, because that limit does not exist.
$endgroup$
– JnxF
Sep 5 '16 at 3:32
5
$begingroup$
You need to consider the domain over which the function is defined. The function $f:[0,infty)rightarrow mathbb{R}$ given by $f(x) = sqrt{x}$ is continuous.
$endgroup$
– Michael
Sep 5 '16 at 3:34
add a comment |
$begingroup$
If a function has a limit from the right but not from the left, is it still continuous?
calculus
asked Sep 5 '16 at 3:28
Rodrigo StvRodrigo Stv
10317
$endgroup$
If a function has a limit from the right but not from the left, is it still continuous?
calculus
calculus
asked Sep 5 '16 at 3:28
Rodrigo StvRodrigo Stv
10317
asked Sep 5 '16 at 3:28
Rodrigo StvRodrigo Stv
10317
edited Sep 5 '16 at 6:44
user99914
asked Sep 5 '16 at 3:28
Rodrigo StvRodrigo Stv
10317
asked Sep 5 '16 at 3:28
Rodrigo StvRodrigo Stv
10317
asked Sep 5 '16 at 3:28
Rodrigo StvRodrigo Stv
10317
10317
$begingroup$
No, because that limit does not exist.
$endgroup$
– JnxF
Sep 5 '16 at 3:32
5
$begingroup$
You need to consider the domain over which the function is defined. The function $f:[0,infty)rightarrow mathbb{R}$ given by $f(x) = sqrt{x}$ is continuous.
$endgroup$
– Michael
Sep 5 '16 at 3:34
add a comment |
$begingroup$
No, because that limit does not exist.
$endgroup$
– JnxF
Sep 5 '16 at 3:32
5
$begingroup$
You need to consider the domain over which the function is defined. The function $f:[0,infty)rightarrow mathbb{R}$ given by $f(x) = sqrt{x}$ is continuous.
$endgroup$
– Michael
Sep 5 '16 at 3:34
$begingroup$
No, because that limit does not exist.
$endgroup$
– JnxF
Sep 5 '16 at 3:32
$begingroup$
No, because that limit does not exist.
$endgroup$
– JnxF
Sep 5 '16 at 3:32
5
5
$begingroup$
You need to consider the domain over which the function is defined. The function $f:[0,infty)rightarrow mathbb{R}$ given by $f(x) = sqrt{x}$ is continuous.
$endgroup$
– Michael
Sep 5 '16 at 3:34
$begingroup$
You need to consider the domain over which the function is defined. The function $f:[0,infty)rightarrow mathbb{R}$ given by $f(x) = sqrt{x}$ is continuous.
$endgroup$
– Michael
Sep 5 '16 at 3:34
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is continuous at $0$.
By construction, the domain of the square-root function is $mathbb R_+=[0,infty)$. Now, for any sequence $(x_n)_{ninmathbb N}$ in the domain (that is, $x_ngeq 0$ for all $ninmathbb N$) that converges to $0$, one has that the corresponding function values $sqrt{x_n}$ also converge to $sqrt{0}=0$.
And this is all you need for continuity by (one of the multiple equivalent) definition(s) of this concept in general metric spaces. What goes on “from the left” is outside of the domain and hence outside of interest as far as continuity is concerned.
answered Sep 5 '16 at 3:37
triple_sectriple_sec
16k21852
$endgroup$
$begingroup$
Thanks for this. It must always be stressed that continuity can be considered only for points in the domain of the function. Failure to do this results in the (false) statement that $f(x)=1/x$ is discontinuous at $0$, a common error of many texts.
$endgroup$
– Lubin
Jan 23 at 16:32
add a comment |
$begingroup$
One can only talk about continuity on the domain where the function exists. The function doesn't exist for $x < 0$ so we can't talk about limits for values $x < 0$.
$sqrt x $ is continuous everywhere it exists.
answered Sep 5 '16 at 3:41
fleabloodfleablood
72.3k22687
$endgroup$
$begingroup$
I assume you meant $x<0$?
$endgroup$
– 3x89g2
Sep 5 '16 at 3:48
$begingroup$
Well, $0$ is in the domain of $sqrt x$, so...? It is absolutely valid to say that a function $f:[0,infty) to mathbb R$ is continuous (or not) at $0$.
$endgroup$
– Pedro Tamaroff♦
Sep 5 '16 at 4:10
1
$begingroup$
So it's continuous at 0.
$endgroup$
– fleablood
Sep 5 '16 at 4:12
$begingroup$
Lots of downvotes for an obvious typo. How rude!
$endgroup$
– MathematicsStudent1122
Sep 5 '16 at 6:54
$begingroup$
@MathematicsStudent1122: The problem wasn't $le$ vs. $<$, but the other wrong statement “we can't talk about continuity at $x=0$” which has now been corrected.
$endgroup$
– Hans Lundmark
Sep 7 '16 at 19:47
|
show 1 more comment
$begingroup$
By definition if the approaches differ, or if one doesn't exist, the limit is undefined, so the function can't be continuous. In this case, however, the limit does exist if you treat it as a function from the complex numbers to the complex numbers, with it approaching zero in the neighborhood of the origin.
answered Sep 5 '16 at 3:32
user361424user361424
1,315414
$endgroup$
$begingroup$
So, @user361424, restricting it to real numbers would make it discontinuous at x = 0?
$endgroup$
– Rodrigo Stv
Sep 5 '16 at 3:33
1
$begingroup$
Not to mention it's simply wrong. Sqrt is continuous everywhere it is defined. At x le 0 it simply doesn't make sense to talk of sqrt being continuous.
$endgroup$
– fleablood
Sep 5 '16 at 3:37
$begingroup$
Yes, restricting it to real numbers would make it discontinuous at 0.
$endgroup$
– user361424
Sep 5 '16 at 3:41
1
$begingroup$
No, it wouldn't.
$endgroup$
– fleablood
Sep 5 '16 at 3:50
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is continuous at $0$.
By construction, the domain of the square-root function is $mathbb R_+=[0,infty)$. Now, for any sequence $(x_n)_{ninmathbb N}$ in the domain (that is, $x_ngeq 0$ for all $ninmathbb N$) that converges to $0$, one has that the corresponding function values $sqrt{x_n}$ also converge to $sqrt{0}=0$.
And this is all you need for continuity by (one of the multiple equivalent) definition(s) of this concept in general metric spaces. What goes on “from the left” is outside of the domain and hence outside of interest as far as continuity is concerned.
answered Sep 5 '16 at 3:37
triple_sectriple_sec
16k21852
$endgroup$
$begingroup$
Thanks for this. It must always be stressed that continuity can be considered only for points in the domain of the function. Failure to do this results in the (false) statement that $f(x)=1/x$ is discontinuous at $0$, a common error of many texts.
$endgroup$
– Lubin
Jan 23 at 16:32
add a comment |
$begingroup$
It is continuous at $0$.
By construction, the domain of the square-root function is $mathbb R_+=[0,infty)$. Now, for any sequence $(x_n)_{ninmathbb N}$ in the domain (that is, $x_ngeq 0$ for all $ninmathbb N$) that converges to $0$, one has that the corresponding function values $sqrt{x_n}$ also converge to $sqrt{0}=0$.
And this is all you need for continuity by (one of the multiple equivalent) definition(s) of this concept in general metric spaces. What goes on “from the left” is outside of the domain and hence outside of interest as far as continuity is concerned.
answered Sep 5 '16 at 3:37
triple_sectriple_sec
16k21852
$endgroup$
$begingroup$
Thanks for this. It must always be stressed that continuity can be considered only for points in the domain of the function. Failure to do this results in the (false) statement that $f(x)=1/x$ is discontinuous at $0$, a common error of many texts.
$endgroup$
– Lubin
Jan 23 at 16:32
add a comment |
$begingroup$
It is continuous at $0$.
By construction, the domain of the square-root function is $mathbb R_+=[0,infty)$. Now, for any sequence $(x_n)_{ninmathbb N}$ in the domain (that is, $x_ngeq 0$ for all $ninmathbb N$) that converges to $0$, one has that the corresponding function values $sqrt{x_n}$ also converge to $sqrt{0}=0$.
And this is all you need for continuity by (one of the multiple equivalent) definition(s) of this concept in general metric spaces. What goes on “from the left” is outside of the domain and hence outside of interest as far as continuity is concerned.
answered Sep 5 '16 at 3:37
triple_sectriple_sec
16k21852
$endgroup$
It is continuous at $0$.
By construction, the domain of the square-root function is $mathbb R_+=[0,infty)$. Now, for any sequence $(x_n)_{ninmathbb N}$ in the domain (that is, $x_ngeq 0$ for all $ninmathbb N$) that converges to $0$, one has that the corresponding function values $sqrt{x_n}$ also converge to $sqrt{0}=0$.
And this is all you need for continuity by (one of the multiple equivalent) definition(s) of this concept in general metric spaces. What goes on “from the left” is outside of the domain and hence outside of interest as far as continuity is concerned.
answered Sep 5 '16 at 3:37
triple_sectriple_sec
16k21852
edited Sep 5 '16 at 6:37
answered Sep 5 '16 at 3:37
triple_sectriple_sec
16k21852
answered Sep 5 '16 at 3:37
triple_sectriple_sec
16k21852
answered Sep 5 '16 at 3:37
triple_sectriple_sec
16k21852
16k21852
$begingroup$
Thanks for this. It must always be stressed that continuity can be considered only for points in the domain of the function. Failure to do this results in the (false) statement that $f(x)=1/x$ is discontinuous at $0$, a common error of many texts.
$endgroup$
– Lubin
Jan 23 at 16:32
add a comment |
$begingroup$
Thanks for this. It must always be stressed that continuity can be considered only for points in the domain of the function. Failure to do this results in the (false) statement that $f(x)=1/x$ is discontinuous at $0$, a common error of many texts.
$endgroup$
– Lubin
Jan 23 at 16:32
$begingroup$
Thanks for this. It must always be stressed that continuity can be considered only for points in the domain of the function. Failure to do this results in the (false) statement that $f(x)=1/x$ is discontinuous at $0$, a common error of many texts.
$endgroup$
– Lubin
Jan 23 at 16:32
$begingroup$
Thanks for this. It must always be stressed that continuity can be considered only for points in the domain of the function. Failure to do this results in the (false) statement that $f(x)=1/x$ is discontinuous at $0$, a common error of many texts.
$endgroup$
– Lubin
Jan 23 at 16:32
add a comment |
$begingroup$
One can only talk about continuity on the domain where the function exists. The function doesn't exist for $x < 0$ so we can't talk about limits for values $x < 0$.
$sqrt x $ is continuous everywhere it exists.
answered Sep 5 '16 at 3:41
fleabloodfleablood
72.3k22687
$endgroup$
$begingroup$
I assume you meant $x<0$?
$endgroup$
– 3x89g2
Sep 5 '16 at 3:48
$begingroup$
Well, $0$ is in the domain of $sqrt x$, so...? It is absolutely valid to say that a function $f:[0,infty) to mathbb R$ is continuous (or not) at $0$.
$endgroup$
– Pedro Tamaroff♦
Sep 5 '16 at 4:10
1
$begingroup$
So it's continuous at 0.
$endgroup$
– fleablood
Sep 5 '16 at 4:12
$begingroup$
Lots of downvotes for an obvious typo. How rude!
$endgroup$
– MathematicsStudent1122
Sep 5 '16 at 6:54
$begingroup$
@MathematicsStudent1122: The problem wasn't $le$ vs. $<$, but the other wrong statement “we can't talk about continuity at $x=0$” which has now been corrected.
$endgroup$
– Hans Lundmark
Sep 7 '16 at 19:47
|
show 1 more comment
$begingroup$
One can only talk about continuity on the domain where the function exists. The function doesn't exist for $x < 0$ so we can't talk about limits for values $x < 0$.
$sqrt x $ is continuous everywhere it exists.
answered Sep 5 '16 at 3:41
fleabloodfleablood
72.3k22687
$endgroup$
$begingroup$
I assume you meant $x<0$?
$endgroup$
– 3x89g2
Sep 5 '16 at 3:48
$begingroup$
Well, $0$ is in the domain of $sqrt x$, so...? It is absolutely valid to say that a function $f:[0,infty) to mathbb R$ is continuous (or not) at $0$.
$endgroup$
– Pedro Tamaroff♦
Sep 5 '16 at 4:10
1
$begingroup$
So it's continuous at 0.
$endgroup$
– fleablood
Sep 5 '16 at 4:12
$begingroup$
Lots of downvotes for an obvious typo. How rude!
$endgroup$
– MathematicsStudent1122
Sep 5 '16 at 6:54
$begingroup$
@MathematicsStudent1122: The problem wasn't $le$ vs. $<$, but the other wrong statement “we can't talk about continuity at $x=0$” which has now been corrected.
$endgroup$
– Hans Lundmark
Sep 7 '16 at 19:47
|
show 1 more comment
$begingroup$
One can only talk about continuity on the domain where the function exists. The function doesn't exist for $x < 0$ so we can't talk about limits for values $x < 0$.
$sqrt x $ is continuous everywhere it exists.
answered Sep 5 '16 at 3:41
fleabloodfleablood
72.3k22687
$endgroup$
One can only talk about continuity on the domain where the function exists. The function doesn't exist for $x < 0$ so we can't talk about limits for values $x < 0$.
$sqrt x $ is continuous everywhere it exists.
answered Sep 5 '16 at 3:41
fleabloodfleablood
72.3k22687
edited Sep 5 '16 at 18:13
answered Sep 5 '16 at 3:41
fleabloodfleablood
72.3k22687
answered Sep 5 '16 at 3:41
fleabloodfleablood
72.3k22687
answered Sep 5 '16 at 3:41
fleabloodfleablood
72.3k22687
72.3k22687
$begingroup$
I assume you meant $x<0$?
$endgroup$
– 3x89g2
Sep 5 '16 at 3:48
$begingroup$
Well, $0$ is in the domain of $sqrt x$, so...? It is absolutely valid to say that a function $f:[0,infty) to mathbb R$ is continuous (or not) at $0$.
$endgroup$
– Pedro Tamaroff♦
Sep 5 '16 at 4:10
1
$begingroup$
So it's continuous at 0.
$endgroup$
– fleablood
Sep 5 '16 at 4:12
$begingroup$
Lots of downvotes for an obvious typo. How rude!
$endgroup$
– MathematicsStudent1122
Sep 5 '16 at 6:54
$begingroup$
@MathematicsStudent1122: The problem wasn't $le$ vs. $<$, but the other wrong statement “we can't talk about continuity at $x=0$” which has now been corrected.
$endgroup$
– Hans Lundmark
Sep 7 '16 at 19:47
|
show 1 more comment
$begingroup$
I assume you meant $x<0$?
$endgroup$
– 3x89g2
Sep 5 '16 at 3:48
$begingroup$
Well, $0$ is in the domain of $sqrt x$, so...? It is absolutely valid to say that a function $f:[0,infty) to mathbb R$ is continuous (or not) at $0$.
$endgroup$
– Pedro Tamaroff♦
Sep 5 '16 at 4:10
1
$begingroup$
So it's continuous at 0.
$endgroup$
– fleablood
Sep 5 '16 at 4:12
$begingroup$
Lots of downvotes for an obvious typo. How rude!
$endgroup$
– MathematicsStudent1122
Sep 5 '16 at 6:54
$begingroup$
@MathematicsStudent1122: The problem wasn't $le$ vs. $<$, but the other wrong statement “we can't talk about continuity at $x=0$” which has now been corrected.
$endgroup$
– Hans Lundmark
Sep 7 '16 at 19:47
$begingroup$
I assume you meant $x<0$?
$endgroup$
– 3x89g2
Sep 5 '16 at 3:48
$begingroup$
I assume you meant $x<0$?
$endgroup$
– 3x89g2
Sep 5 '16 at 3:48
$begingroup$
Well, $0$ is in the domain of $sqrt x$, so...? It is absolutely valid to say that a function $f:[0,infty) to mathbb R$ is continuous (or not) at $0$.
$endgroup$
– Pedro Tamaroff♦
Sep 5 '16 at 4:10
$begingroup$
Well, $0$ is in the domain of $sqrt x$, so...? It is absolutely valid to say that a function $f:[0,infty) to mathbb R$ is continuous (or not) at $0$.
$endgroup$
– Pedro Tamaroff♦
Sep 5 '16 at 4:10
1
1
$begingroup$
So it's continuous at 0.
$endgroup$
– fleablood
Sep 5 '16 at 4:12
$begingroup$
So it's continuous at 0.
$endgroup$
– fleablood
Sep 5 '16 at 4:12
$begingroup$
Lots of downvotes for an obvious typo. How rude!
$endgroup$
– MathematicsStudent1122
Sep 5 '16 at 6:54
$begingroup$
Lots of downvotes for an obvious typo. How rude!
$endgroup$
– MathematicsStudent1122
Sep 5 '16 at 6:54
$begingroup$
@MathematicsStudent1122: The problem wasn't $le$ vs. $<$, but the other wrong statement “we can't talk about continuity at $x=0$” which has now been corrected.
$endgroup$
– Hans Lundmark
Sep 7 '16 at 19:47
$begingroup$
@MathematicsStudent1122: The problem wasn't $le$ vs. $<$, but the other wrong statement “we can't talk about continuity at $x=0$” which has now been corrected.
$endgroup$
– Hans Lundmark
Sep 7 '16 at 19:47
|
show 1 more comment
$begingroup$
By definition if the approaches differ, or if one doesn't exist, the limit is undefined, so the function can't be continuous. In this case, however, the limit does exist if you treat it as a function from the complex numbers to the complex numbers, with it approaching zero in the neighborhood of the origin.
answered Sep 5 '16 at 3:32
user361424user361424
1,315414
$endgroup$
$begingroup$
So, @user361424, restricting it to real numbers would make it discontinuous at x = 0?
$endgroup$
– Rodrigo Stv
Sep 5 '16 at 3:33
1
$begingroup$
Not to mention it's simply wrong. Sqrt is continuous everywhere it is defined. At x le 0 it simply doesn't make sense to talk of sqrt being continuous.
$endgroup$
– fleablood
Sep 5 '16 at 3:37
$begingroup$
Yes, restricting it to real numbers would make it discontinuous at 0.
$endgroup$
– user361424
Sep 5 '16 at 3:41
1
$begingroup$
No, it wouldn't.
$endgroup$
– fleablood
Sep 5 '16 at 3:50
add a comment |
$begingroup$
By definition if the approaches differ, or if one doesn't exist, the limit is undefined, so the function can't be continuous. In this case, however, the limit does exist if you treat it as a function from the complex numbers to the complex numbers, with it approaching zero in the neighborhood of the origin.
answered Sep 5 '16 at 3:32
user361424user361424
1,315414
$endgroup$
$begingroup$
So, @user361424, restricting it to real numbers would make it discontinuous at x = 0?
$endgroup$
– Rodrigo Stv
Sep 5 '16 at 3:33
1
$begingroup$
Not to mention it's simply wrong. Sqrt is continuous everywhere it is defined. At x le 0 it simply doesn't make sense to talk of sqrt being continuous.
$endgroup$
– fleablood
Sep 5 '16 at 3:37
$begingroup$
Yes, restricting it to real numbers would make it discontinuous at 0.
$endgroup$
– user361424
Sep 5 '16 at 3:41
1
$begingroup$
No, it wouldn't.
$endgroup$
– fleablood
Sep 5 '16 at 3:50
add a comment |
$begingroup$
By definition if the approaches differ, or if one doesn't exist, the limit is undefined, so the function can't be continuous. In this case, however, the limit does exist if you treat it as a function from the complex numbers to the complex numbers, with it approaching zero in the neighborhood of the origin.
answered Sep 5 '16 at 3:32
user361424user361424
1,315414
$endgroup$
By definition if the approaches differ, or if one doesn't exist, the limit is undefined, so the function can't be continuous. In this case, however, the limit does exist if you treat it as a function from the complex numbers to the complex numbers, with it approaching zero in the neighborhood of the origin.
answered Sep 5 '16 at 3:32
user361424user361424
1,315414
answered Sep 5 '16 at 3:32
user361424user361424
1,315414
answered Sep 5 '16 at 3:32
user361424user361424
1,315414
answered Sep 5 '16 at 3:32
user361424user361424
1,315414
1,315414
$begingroup$
So, @user361424, restricting it to real numbers would make it discontinuous at x = 0?
$endgroup$
– Rodrigo Stv
Sep 5 '16 at 3:33
1
$begingroup$
Not to mention it's simply wrong. Sqrt is continuous everywhere it is defined. At x le 0 it simply doesn't make sense to talk of sqrt being continuous.
$endgroup$
– fleablood
Sep 5 '16 at 3:37
$begingroup$
Yes, restricting it to real numbers would make it discontinuous at 0.
$endgroup$
– user361424
Sep 5 '16 at 3:41
1
$begingroup$
No, it wouldn't.
$endgroup$
– fleablood
Sep 5 '16 at 3:50
add a comment |
$begingroup$
So, @user361424, restricting it to real numbers would make it discontinuous at x = 0?
$endgroup$
– Rodrigo Stv
Sep 5 '16 at 3:33
1
$begingroup$
Not to mention it's simply wrong. Sqrt is continuous everywhere it is defined. At x le 0 it simply doesn't make sense to talk of sqrt being continuous.
$endgroup$
– fleablood
Sep 5 '16 at 3:37
$begingroup$
Yes, restricting it to real numbers would make it discontinuous at 0.
$endgroup$
– user361424
Sep 5 '16 at 3:41
1
$begingroup$
No, it wouldn't.
$endgroup$
– fleablood
Sep 5 '16 at 3:50
$begingroup$
So, @user361424, restricting it to real numbers would make it discontinuous at x = 0?
$endgroup$
– Rodrigo Stv
Sep 5 '16 at 3:33
$begingroup$
So, @user361424, restricting it to real numbers would make it discontinuous at x = 0?
$endgroup$
– Rodrigo Stv
Sep 5 '16 at 3:33
1
1
$begingroup$
Not to mention it's simply wrong. Sqrt is continuous everywhere it is defined. At x le 0 it simply doesn't make sense to talk of sqrt being continuous.
$endgroup$
– fleablood
Sep 5 '16 at 3:37
$begingroup$
Not to mention it's simply wrong. Sqrt is continuous everywhere it is defined. At x le 0 it simply doesn't make sense to talk of sqrt being continuous.
$endgroup$
– fleablood
Sep 5 '16 at 3:37
$begingroup$
Yes, restricting it to real numbers would make it discontinuous at 0.
$endgroup$
– user361424
Sep 5 '16 at 3:41
$begingroup$
Yes, restricting it to real numbers would make it discontinuous at 0.
$endgroup$
– user361424
Sep 5 '16 at 3:41
1
1
$begingroup$
No, it wouldn't.
$endgroup$
– fleablood
Sep 5 '16 at 3:50
$begingroup$
No, it wouldn't.
$endgroup$
– fleablood
Sep 5 '16 at 3:50
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$begingroup$
No, because that limit does not exist.
$endgroup$
– JnxF
Sep 5 '16 at 3:32
5
$begingroup$
You need to consider the domain over which the function is defined. The function $f:[0,infty)rightarrow mathbb{R}$ given by $f(x) = sqrt{x}$ is continuous.
$endgroup$
– Michael
Sep 5 '16 at 3:34