Prove that if $f:(a, b) to mathbb{R}$ is defined on an open interval then $f$ is continuous iff $ lim_{xto c}...
If $$f(x) = begin{cases}18-x^2 &x<2 \x^2+2x+7 &xgeq2 end{cases}$$
Find a sequence such that $x_nto2$ and $f(x_n)notto f(2)$
I found a definition that says
$f:(a, b) to mathbb{R}$ is defined on an open interval then $f$ is continuous iff $displaystylelim_{xto c} f(x) =f(c): text{for}: a < c < b$ since every point is an accumulation point.
So I know that it there is a function where $x_nto2$ and $f(x_n) notto f(2)$ since the $f(x)$ is not continuous at 2 but I have no idea what this function would be.
As an attempt I tried $x_n=(2+frac1n)$ since its limit is $2$ but this did not look correct.
Any help appreciated
real-analysis continuity
asked Nov 23 at 17:54
James
154
closed as unclear what you're asking by Andrés E. Caicedo, Leucippus, Cesareo, ancientmathematician, Rebellos Nov 24 at 9:08
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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If $$f(x) = begin{cases}18-x^2 &x<2 \x^2+2x+7 &xgeq2 end{cases}$$
Find a sequence such that $x_nto2$ and $f(x_n)notto f(2)$
I found a definition that says
$f:(a, b) to mathbb{R}$ is defined on an open interval then $f$ is continuous iff $displaystylelim_{xto c} f(x) =f(c): text{for}: a < c < b$ since every point is an accumulation point.
So I know that it there is a function where $x_nto2$ and $f(x_n) notto f(2)$ since the $f(x)$ is not continuous at 2 but I have no idea what this function would be.
As an attempt I tried $x_n=(2+frac1n)$ since its limit is $2$ but this did not look correct.
Any help appreciated
real-analysis continuity
asked Nov 23 at 17:54
James
154
closed as unclear what you're asking by Andrés E. Caicedo, Leucippus, Cesareo, ancientmathematician, Rebellos Nov 24 at 9:08
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
refer math.meta.stackexchange.com/questions/5020/… for proper formatting. I've done it for you this time.
– idea
Nov 23 at 18:01
2
What is $f(2)$.
– hamam_Abdallah
Nov 23 at 18:49
add a comment |
If $$f(x) = begin{cases}18-x^2 &x<2 \x^2+2x+7 &xgeq2 end{cases}$$
Find a sequence such that $x_nto2$ and $f(x_n)notto f(2)$
I found a definition that says
$f:(a, b) to mathbb{R}$ is defined on an open interval then $f$ is continuous iff $displaystylelim_{xto c} f(x) =f(c): text{for}: a < c < b$ since every point is an accumulation point.
So I know that it there is a function where $x_nto2$ and $f(x_n) notto f(2)$ since the $f(x)$ is not continuous at 2 but I have no idea what this function would be.
As an attempt I tried $x_n=(2+frac1n)$ since its limit is $2$ but this did not look correct.
Any help appreciated
real-analysis continuity
asked Nov 23 at 17:54
James
154
If $$f(x) = begin{cases}18-x^2 &x<2 \x^2+2x+7 &xgeq2 end{cases}$$
Find a sequence such that $x_nto2$ and $f(x_n)notto f(2)$
I found a definition that says
$f:(a, b) to mathbb{R}$ is defined on an open interval then $f$ is continuous iff $displaystylelim_{xto c} f(x) =f(c): text{for}: a < c < b$ since every point is an accumulation point.
So I know that it there is a function where $x_nto2$ and $f(x_n) notto f(2)$ since the $f(x)$ is not continuous at 2 but I have no idea what this function would be.
As an attempt I tried $x_n=(2+frac1n)$ since its limit is $2$ but this did not look correct.
Any help appreciated
real-analysis continuity
real-analysis continuity
asked Nov 23 at 17:54
James
154
asked Nov 23 at 17:54
James
154
edited Nov 24 at 11:32
asked Nov 23 at 17:54
James
154
asked Nov 23 at 17:54
James
154
asked Nov 23 at 17:54
James
154
154
closed as unclear what you're asking by Andrés E. Caicedo, Leucippus, Cesareo, ancientmathematician, Rebellos Nov 24 at 9:08
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Andrés E. Caicedo, Leucippus, Cesareo, ancientmathematician, Rebellos Nov 24 at 9:08
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
refer math.meta.stackexchange.com/questions/5020/… for proper formatting. I've done it for you this time.
– idea
Nov 23 at 18:01
2
What is $f(2)$.
– hamam_Abdallah
Nov 23 at 18:49
add a comment |
refer math.meta.stackexchange.com/questions/5020/… for proper formatting. I've done it for you this time.
– idea
Nov 23 at 18:01
2
What is $f(2)$.
– hamam_Abdallah
Nov 23 at 18:49
refer math.meta.stackexchange.com/questions/5020/… for proper formatting. I've done it for you this time.
– idea
Nov 23 at 18:01
refer math.meta.stackexchange.com/questions/5020/… for proper formatting. I've done it for you this time.
– idea
Nov 23 at 18:01
2
2
What is $f(2)$.
– hamam_Abdallah
Nov 23 at 18:49
What is $f(2)$.
– hamam_Abdallah
Nov 23 at 18:49
add a comment |
2 Answers
2
active
oldest
votes
Let $x_{n}:=2-frac{1}{n}$. Then, for each $n$, $x_{n}< 2$, so
$$
f(x_{n})=18-left(2-frac{1}{n}right)^{2}=18-4+frac{4}{n}-frac{1}{n^{2}}to 14qquadtext{as }ntoinfty.
$$
But, $f(2)=15$.
answered Nov 23 at 17:57
ervx
10.3k31338
Appreciate the help!
– James
Nov 23 at 18:01
add a comment |
You can take
$$x_n=2+frac{(-1)^n}{n+1}$$
$$(x_n)to 2$$
$$x_{2n}>2implies f(x_{2n})to 15$$
$$x_{2n+1}<2implies f(x_{2n+1})to 14$$
answered Nov 23 at 18:45
hamam_Abdallah
37.9k21634
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $x_{n}:=2-frac{1}{n}$. Then, for each $n$, $x_{n}< 2$, so
$$
f(x_{n})=18-left(2-frac{1}{n}right)^{2}=18-4+frac{4}{n}-frac{1}{n^{2}}to 14qquadtext{as }ntoinfty.
$$
But, $f(2)=15$.
answered Nov 23 at 17:57
ervx
10.3k31338
Appreciate the help!
– James
Nov 23 at 18:01
add a comment |
Let $x_{n}:=2-frac{1}{n}$. Then, for each $n$, $x_{n}< 2$, so
$$
f(x_{n})=18-left(2-frac{1}{n}right)^{2}=18-4+frac{4}{n}-frac{1}{n^{2}}to 14qquadtext{as }ntoinfty.
$$
But, $f(2)=15$.
answered Nov 23 at 17:57
ervx
10.3k31338
Appreciate the help!
– James
Nov 23 at 18:01
add a comment |
Let $x_{n}:=2-frac{1}{n}$. Then, for each $n$, $x_{n}< 2$, so
$$
f(x_{n})=18-left(2-frac{1}{n}right)^{2}=18-4+frac{4}{n}-frac{1}{n^{2}}to 14qquadtext{as }ntoinfty.
$$
But, $f(2)=15$.
answered Nov 23 at 17:57
ervx
10.3k31338
Let $x_{n}:=2-frac{1}{n}$. Then, for each $n$, $x_{n}< 2$, so
$$
f(x_{n})=18-left(2-frac{1}{n}right)^{2}=18-4+frac{4}{n}-frac{1}{n^{2}}to 14qquadtext{as }ntoinfty.
$$
But, $f(2)=15$.
answered Nov 23 at 17:57
ervx
10.3k31338
answered Nov 23 at 17:57
ervx
10.3k31338
answered Nov 23 at 17:57
ervx
10.3k31338
answered Nov 23 at 17:57
ervx
10.3k31338
10.3k31338
Appreciate the help!
– James
Nov 23 at 18:01
add a comment |
Appreciate the help!
– James
Nov 23 at 18:01
Appreciate the help!
– James
Nov 23 at 18:01
Appreciate the help!
– James
Nov 23 at 18:01
add a comment |
You can take
$$x_n=2+frac{(-1)^n}{n+1}$$
$$(x_n)to 2$$
$$x_{2n}>2implies f(x_{2n})to 15$$
$$x_{2n+1}<2implies f(x_{2n+1})to 14$$
answered Nov 23 at 18:45
hamam_Abdallah
37.9k21634
add a comment |
You can take
$$x_n=2+frac{(-1)^n}{n+1}$$
$$(x_n)to 2$$
$$x_{2n}>2implies f(x_{2n})to 15$$
$$x_{2n+1}<2implies f(x_{2n+1})to 14$$
answered Nov 23 at 18:45
hamam_Abdallah
37.9k21634
add a comment |
You can take
$$x_n=2+frac{(-1)^n}{n+1}$$
$$(x_n)to 2$$
$$x_{2n}>2implies f(x_{2n})to 15$$
$$x_{2n+1}<2implies f(x_{2n+1})to 14$$
answered Nov 23 at 18:45
hamam_Abdallah
37.9k21634
You can take
$$x_n=2+frac{(-1)^n}{n+1}$$
$$(x_n)to 2$$
$$x_{2n}>2implies f(x_{2n})to 15$$
$$x_{2n+1}<2implies f(x_{2n+1})to 14$$
answered Nov 23 at 18:45
hamam_Abdallah
37.9k21634
answered Nov 23 at 18:45
hamam_Abdallah
37.9k21634
answered Nov 23 at 18:45
hamam_Abdallah
37.9k21634
answered Nov 23 at 18:45
hamam_Abdallah
37.9k21634
37.9k21634
add a comment |
add a comment |
refer math.meta.stackexchange.com/questions/5020/… for proper formatting. I've done it for you this time.
– idea
Nov 23 at 18:01
2
What is $f(2)$.
– hamam_Abdallah
Nov 23 at 18:49