Mixing
How to prove that $sqrt x$ is continuous in $[0,infty)$?
$begingroup$
I am trying to prove that
$sqrt x$ is continuous in $[0,infty)$.
I have started writing the following proof:
Given $x_0 in [0,infty)$ and $epsilon > 0$. We have to show that there exists a $delta > 0$ such that for every $x in (x_0 - delta,x_0 +delta)$, $sqrt x in (sqrt x_0 - epsilon, sqrt x_0 + epsilon)$.
So, $| sqrt x - sqrt x_0 | = frac {| (sqrt x - sqrt x_0)(sqrt x + sqrt x_0) |}{| sqrt x + sqrt x_0 |} = frac {| x - x_0 |}{| sqrt x + sqrt x_0 |}$
I am not sure how to continue from here..
I could take $M = max{ x,x_0 }$ and $delta = 2M epsilon$ but this is unnecessary if $x,x_0 > 1$.
Should I split to cases? $x,x_0 > 1$, $0 < x,x_0 < 1$ etc..
Should I dea with $x_0 = 0$ separately with a right neighborhood $[0,delta)$?
I feel like I am over complicating this..
What is the simplest way to define $delta$?
Thanks!!
real-analysis calculus complex-analysis continuity uniform-continuity
asked Jan 8 at 11:40
user135172user135172
43629
$endgroup$
add a comment |
$begingroup$
I am trying to prove that
$sqrt x$ is continuous in $[0,infty)$.
I have started writing the following proof:
Given $x_0 in [0,infty)$ and $epsilon > 0$. We have to show that there exists a $delta > 0$ such that for every $x in (x_0 - delta,x_0 +delta)$, $sqrt x in (sqrt x_0 - epsilon, sqrt x_0 + epsilon)$.
So, $| sqrt x - sqrt x_0 | = frac {| (sqrt x - sqrt x_0)(sqrt x + sqrt x_0) |}{| sqrt x + sqrt x_0 |} = frac {| x - x_0 |}{| sqrt x + sqrt x_0 |}$
I am not sure how to continue from here..
I could take $M = max{ x,x_0 }$ and $delta = 2M epsilon$ but this is unnecessary if $x,x_0 > 1$.
Should I split to cases? $x,x_0 > 1$, $0 < x,x_0 < 1$ etc..
Should I dea with $x_0 = 0$ separately with a right neighborhood $[0,delta)$?
I feel like I am over complicating this..
What is the simplest way to define $delta$?
Thanks!!
real-analysis calculus complex-analysis continuity uniform-continuity
asked Jan 8 at 11:40
user135172user135172
43629
$endgroup$
$begingroup$
Various solutions at math.stackexchange.com/q/560307/42969.
$endgroup$
– Martin R
Jan 8 at 12:02
$begingroup$
Your reasoning is correct, technically it translates into $M:=min(max(x_0,x),1)$. Unnecessary doesn't make it bad. To make it really simple, you should use the fact that this is the inverse of $xmapsto x^2$ on $[0,infty)$ which is a continuous and monotonic function (a homeomorphism in fancy language), which must have a continuous and monotonic inverse.
$endgroup$
– Oskar Limka
Jan 8 at 12:13
$begingroup$
Also related: math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Jan 8 at 12:13
add a comment |
$begingroup$
I am trying to prove that
$sqrt x$ is continuous in $[0,infty)$.
I have started writing the following proof:
Given $x_0 in [0,infty)$ and $epsilon > 0$. We have to show that there exists a $delta > 0$ such that for every $x in (x_0 - delta,x_0 +delta)$, $sqrt x in (sqrt x_0 - epsilon, sqrt x_0 + epsilon)$.
So, $| sqrt x - sqrt x_0 | = frac {| (sqrt x - sqrt x_0)(sqrt x + sqrt x_0) |}{| sqrt x + sqrt x_0 |} = frac {| x - x_0 |}{| sqrt x + sqrt x_0 |}$
I am not sure how to continue from here..
I could take $M = max{ x,x_0 }$ and $delta = 2M epsilon$ but this is unnecessary if $x,x_0 > 1$.
Should I split to cases? $x,x_0 > 1$, $0 < x,x_0 < 1$ etc..
Should I dea with $x_0 = 0$ separately with a right neighborhood $[0,delta)$?
I feel like I am over complicating this..
What is the simplest way to define $delta$?
Thanks!!
real-analysis calculus complex-analysis continuity uniform-continuity
asked Jan 8 at 11:40
user135172user135172
43629
$endgroup$
I am trying to prove that
$sqrt x$ is continuous in $[0,infty)$.
I have started writing the following proof:
Given $x_0 in [0,infty)$ and $epsilon > 0$. We have to show that there exists a $delta > 0$ such that for every $x in (x_0 - delta,x_0 +delta)$, $sqrt x in (sqrt x_0 - epsilon, sqrt x_0 + epsilon)$.
So, $| sqrt x - sqrt x_0 | = frac {| (sqrt x - sqrt x_0)(sqrt x + sqrt x_0) |}{| sqrt x + sqrt x_0 |} = frac {| x - x_0 |}{| sqrt x + sqrt x_0 |}$
I am not sure how to continue from here..
I could take $M = max{ x,x_0 }$ and $delta = 2M epsilon$ but this is unnecessary if $x,x_0 > 1$.
Should I split to cases? $x,x_0 > 1$, $0 < x,x_0 < 1$ etc..
Should I dea with $x_0 = 0$ separately with a right neighborhood $[0,delta)$?
I feel like I am over complicating this..
What is the simplest way to define $delta$?
Thanks!!
real-analysis calculus complex-analysis continuity uniform-continuity
real-analysis calculus complex-analysis continuity uniform-continuity
asked Jan 8 at 11:40
user135172user135172
43629
asked Jan 8 at 11:40
user135172user135172
43629
edited Jan 8 at 11:46
user135172
asked Jan 8 at 11:40
user135172user135172
43629
asked Jan 8 at 11:40
user135172user135172
43629
asked Jan 8 at 11:40
user135172user135172
43629
43629
$begingroup$
Various solutions at math.stackexchange.com/q/560307/42969.
$endgroup$
– Martin R
Jan 8 at 12:02
$begingroup$
Your reasoning is correct, technically it translates into $M:=min(max(x_0,x),1)$. Unnecessary doesn't make it bad. To make it really simple, you should use the fact that this is the inverse of $xmapsto x^2$ on $[0,infty)$ which is a continuous and monotonic function (a homeomorphism in fancy language), which must have a continuous and monotonic inverse.
$endgroup$
– Oskar Limka
Jan 8 at 12:13
$begingroup$
Also related: math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Jan 8 at 12:13
add a comment |
$begingroup$
Various solutions at math.stackexchange.com/q/560307/42969.
$endgroup$
– Martin R
Jan 8 at 12:02
$begingroup$
Your reasoning is correct, technically it translates into $M:=min(max(x_0,x),1)$. Unnecessary doesn't make it bad. To make it really simple, you should use the fact that this is the inverse of $xmapsto x^2$ on $[0,infty)$ which is a continuous and monotonic function (a homeomorphism in fancy language), which must have a continuous and monotonic inverse.
$endgroup$
– Oskar Limka
Jan 8 at 12:13
$begingroup$
Also related: math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Jan 8 at 12:13
$begingroup$
Various solutions at math.stackexchange.com/q/560307/42969.
$endgroup$
– Martin R
Jan 8 at 12:02
$begingroup$
Various solutions at math.stackexchange.com/q/560307/42969.
$endgroup$
– Martin R
Jan 8 at 12:02
$begingroup$
Your reasoning is correct, technically it translates into $M:=min(max(x_0,x),1)$. Unnecessary doesn't make it bad. To make it really simple, you should use the fact that this is the inverse of $xmapsto x^2$ on $[0,infty)$ which is a continuous and monotonic function (a homeomorphism in fancy language), which must have a continuous and monotonic inverse.
$endgroup$
– Oskar Limka
Jan 8 at 12:13
$begingroup$
Your reasoning is correct, technically it translates into $M:=min(max(x_0,x),1)$. Unnecessary doesn't make it bad. To make it really simple, you should use the fact that this is the inverse of $xmapsto x^2$ on $[0,infty)$ which is a continuous and monotonic function (a homeomorphism in fancy language), which must have a continuous and monotonic inverse.
$endgroup$
– Oskar Limka
Jan 8 at 12:13
$begingroup$
Also related: math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Jan 8 at 12:13
$begingroup$
Also related: math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Jan 8 at 12:13
add a comment |
3 Answers
3
active
oldest
votes
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I will prove the stronger result that the function is uniformly continuous. If $x <epsilon ^{2} /4$ and $x_o <epsilon ^{2} /4$ then then $|sqrt x -sqrt {x_0}| <epsilon /2+epsilon/ 2=epsilon$. Otherwise $|sqrt x -sqrt {x_0}| leq frac {2|x-x_0|} {epsilon}$ from the inequality you have derived. Hence $|sqrt x -sqrt {x_0}|<epsilon$ if $|x-x_0| <epsilon^{2}/2$.
answered Jan 8 at 12:04
Kavi Rama MurthyKavi Rama Murthy
57k42159
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add a comment |
$begingroup$
For $x_{0}neq 0$, we have
$$
|sqrt{x}-sqrt{x_{0}}|=frac{|x-x_{0}|}{sqrt{x}+sqrt{x_{0}}}leq frac{|x-x_{0}|}{sqrt{x_{0}}}.
$$
So you can take $delta=sqrt{x_{0}}epsilon$.
For $x_{0}=0$, we need $sqrt{x}<epsilon$ for $xin(0,delta)$, so we can take $delta=epsilon^{2}$.
answered Jan 8 at 12:03
studiosusstudiosus
1,857713
$endgroup$
add a comment |
$begingroup$
Inverse of a continuous function is continuous on the interval hence the $sqrt x$ is continuous off course being the inverse of square function
answered Jan 8 at 12:19
Bijayan RayBijayan Ray
25110
$endgroup$
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
$begingroup$
I will prove the stronger result that the function is uniformly continuous. If $x <epsilon ^{2} /4$ and $x_o <epsilon ^{2} /4$ then then $|sqrt x -sqrt {x_0}| <epsilon /2+epsilon/ 2=epsilon$. Otherwise $|sqrt x -sqrt {x_0}| leq frac {2|x-x_0|} {epsilon}$ from the inequality you have derived. Hence $|sqrt x -sqrt {x_0}|<epsilon$ if $|x-x_0| <epsilon^{2}/2$.
answered Jan 8 at 12:04
Kavi Rama MurthyKavi Rama Murthy
57k42159
$endgroup$
add a comment |
$begingroup$
I will prove the stronger result that the function is uniformly continuous. If $x <epsilon ^{2} /4$ and $x_o <epsilon ^{2} /4$ then then $|sqrt x -sqrt {x_0}| <epsilon /2+epsilon/ 2=epsilon$. Otherwise $|sqrt x -sqrt {x_0}| leq frac {2|x-x_0|} {epsilon}$ from the inequality you have derived. Hence $|sqrt x -sqrt {x_0}|<epsilon$ if $|x-x_0| <epsilon^{2}/2$.
answered Jan 8 at 12:04
Kavi Rama MurthyKavi Rama Murthy
57k42159
$endgroup$
add a comment |
$begingroup$
I will prove the stronger result that the function is uniformly continuous. If $x <epsilon ^{2} /4$ and $x_o <epsilon ^{2} /4$ then then $|sqrt x -sqrt {x_0}| <epsilon /2+epsilon/ 2=epsilon$. Otherwise $|sqrt x -sqrt {x_0}| leq frac {2|x-x_0|} {epsilon}$ from the inequality you have derived. Hence $|sqrt x -sqrt {x_0}|<epsilon$ if $|x-x_0| <epsilon^{2}/2$.
answered Jan 8 at 12:04
Kavi Rama MurthyKavi Rama Murthy
57k42159
$endgroup$
I will prove the stronger result that the function is uniformly continuous. If $x <epsilon ^{2} /4$ and $x_o <epsilon ^{2} /4$ then then $|sqrt x -sqrt {x_0}| <epsilon /2+epsilon/ 2=epsilon$. Otherwise $|sqrt x -sqrt {x_0}| leq frac {2|x-x_0|} {epsilon}$ from the inequality you have derived. Hence $|sqrt x -sqrt {x_0}|<epsilon$ if $|x-x_0| <epsilon^{2}/2$.
answered Jan 8 at 12:04
Kavi Rama MurthyKavi Rama Murthy
57k42159
answered Jan 8 at 12:04
Kavi Rama MurthyKavi Rama Murthy
57k42159
answered Jan 8 at 12:04
Kavi Rama MurthyKavi Rama Murthy
57k42159
answered Jan 8 at 12:04
Kavi Rama MurthyKavi Rama Murthy
57k42159
57k42159
add a comment |
add a comment |
$begingroup$
For $x_{0}neq 0$, we have
$$
|sqrt{x}-sqrt{x_{0}}|=frac{|x-x_{0}|}{sqrt{x}+sqrt{x_{0}}}leq frac{|x-x_{0}|}{sqrt{x_{0}}}.
$$
So you can take $delta=sqrt{x_{0}}epsilon$.
For $x_{0}=0$, we need $sqrt{x}<epsilon$ for $xin(0,delta)$, so we can take $delta=epsilon^{2}$.
answered Jan 8 at 12:03
studiosusstudiosus
1,857713
$endgroup$
add a comment |
$begingroup$
For $x_{0}neq 0$, we have
$$
|sqrt{x}-sqrt{x_{0}}|=frac{|x-x_{0}|}{sqrt{x}+sqrt{x_{0}}}leq frac{|x-x_{0}|}{sqrt{x_{0}}}.
$$
So you can take $delta=sqrt{x_{0}}epsilon$.
For $x_{0}=0$, we need $sqrt{x}<epsilon$ for $xin(0,delta)$, so we can take $delta=epsilon^{2}$.
answered Jan 8 at 12:03
studiosusstudiosus
1,857713
$endgroup$
add a comment |
$begingroup$
For $x_{0}neq 0$, we have
$$
|sqrt{x}-sqrt{x_{0}}|=frac{|x-x_{0}|}{sqrt{x}+sqrt{x_{0}}}leq frac{|x-x_{0}|}{sqrt{x_{0}}}.
$$
So you can take $delta=sqrt{x_{0}}epsilon$.
For $x_{0}=0$, we need $sqrt{x}<epsilon$ for $xin(0,delta)$, so we can take $delta=epsilon^{2}$.
answered Jan 8 at 12:03
studiosusstudiosus
1,857713
$endgroup$
For $x_{0}neq 0$, we have
$$
|sqrt{x}-sqrt{x_{0}}|=frac{|x-x_{0}|}{sqrt{x}+sqrt{x_{0}}}leq frac{|x-x_{0}|}{sqrt{x_{0}}}.
$$
So you can take $delta=sqrt{x_{0}}epsilon$.
For $x_{0}=0$, we need $sqrt{x}<epsilon$ for $xin(0,delta)$, so we can take $delta=epsilon^{2}$.
answered Jan 8 at 12:03
studiosusstudiosus
1,857713
answered Jan 8 at 12:03
studiosusstudiosus
1,857713
answered Jan 8 at 12:03
studiosusstudiosus
1,857713
answered Jan 8 at 12:03
studiosusstudiosus
1,857713
1,857713
add a comment |
add a comment |
$begingroup$
Inverse of a continuous function is continuous on the interval hence the $sqrt x$ is continuous off course being the inverse of square function
answered Jan 8 at 12:19
Bijayan RayBijayan Ray
25110
$endgroup$
add a comment |
$begingroup$
Inverse of a continuous function is continuous on the interval hence the $sqrt x$ is continuous off course being the inverse of square function
answered Jan 8 at 12:19
Bijayan RayBijayan Ray
25110
$endgroup$
add a comment |
$begingroup$
Inverse of a continuous function is continuous on the interval hence the $sqrt x$ is continuous off course being the inverse of square function
answered Jan 8 at 12:19
Bijayan RayBijayan Ray
25110
$endgroup$
Inverse of a continuous function is continuous on the interval hence the $sqrt x$ is continuous off course being the inverse of square function
answered Jan 8 at 12:19
Bijayan RayBijayan Ray
25110
answered Jan 8 at 12:19
Bijayan RayBijayan Ray
25110
answered Jan 8 at 12:19
Bijayan RayBijayan Ray
25110
answered Jan 8 at 12:19
Bijayan RayBijayan Ray
25110
25110
add a comment |
add a comment |
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$begingroup$
Various solutions at math.stackexchange.com/q/560307/42969.
$endgroup$
– Martin R
Jan 8 at 12:02
$begingroup$
Your reasoning is correct, technically it translates into $M:=min(max(x_0,x),1)$. Unnecessary doesn't make it bad. To make it really simple, you should use the fact that this is the inverse of $xmapsto x^2$ on $[0,infty)$ which is a continuous and monotonic function (a homeomorphism in fancy language), which must have a continuous and monotonic inverse.
$endgroup$
– Oskar Limka
Jan 8 at 12:13
$begingroup$
Also related: math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Jan 8 at 12:13