Prove that $ frac{1}{x^5} $ is not uniformly continuous on $(0,2)$
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1
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Prove that $frac{1}{x^5} $ is not uniformly continuous on $(0,2)$. I tried by putting $epsilon = 1$ but can't seem to find $x$ and $x'$ that would work.
calculus
edited Nov 15 at 19:50
Ernie060
2,520319
asked Nov 15 at 19:44
user15269
1608
add a comment |
up vote
1
down vote
favorite
Prove that $frac{1}{x^5} $ is not uniformly continuous on $(0,2)$. I tried by putting $epsilon = 1$ but can't seem to find $x$ and $x'$ that would work.
calculus
edited Nov 15 at 19:50
Ernie060
2,520319
asked Nov 15 at 19:44
user15269
1608
$(0,2)$...................
– Nosrati
Nov 15 at 19:48
3
A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
– Daniel Schepler
Nov 15 at 20:01
This is a very good comment. Thanks a lot.
– hamam_Abdallah
Nov 16 at 18:02
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Prove that $frac{1}{x^5} $ is not uniformly continuous on $(0,2)$. I tried by putting $epsilon = 1$ but can't seem to find $x$ and $x'$ that would work.
calculus
edited Nov 15 at 19:50
Ernie060
2,520319
asked Nov 15 at 19:44
user15269
1608
Prove that $frac{1}{x^5} $ is not uniformly continuous on $(0,2)$. I tried by putting $epsilon = 1$ but can't seem to find $x$ and $x'$ that would work.
calculus
calculus
edited Nov 15 at 19:50
Ernie060
2,520319
asked Nov 15 at 19:44
user15269
1608
edited Nov 15 at 19:50
Ernie060
2,520319
asked Nov 15 at 19:44
user15269
1608
edited Nov 15 at 19:50
Ernie060
2,520319
edited Nov 15 at 19:50
Ernie060
2,520319
edited Nov 15 at 19:50
Ernie060
2,520319
2,520319
asked Nov 15 at 19:44
user15269
1608
asked Nov 15 at 19:44
user15269
1608
asked Nov 15 at 19:44
user15269
1608
1608
$(0,2)$...................
– Nosrati
Nov 15 at 19:48
3
A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
– Daniel Schepler
Nov 15 at 20:01
This is a very good comment. Thanks a lot.
– hamam_Abdallah
Nov 16 at 18:02
add a comment |
$(0,2)$...................
– Nosrati
Nov 15 at 19:48
3
A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
– Daniel Schepler
Nov 15 at 20:01
This is a very good comment. Thanks a lot.
– hamam_Abdallah
Nov 16 at 18:02
$(0,2)$...................
– Nosrati
Nov 15 at 19:48
$(0,2)$...................
– Nosrati
Nov 15 at 19:48
3
3
A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
– Daniel Schepler
Nov 15 at 20:01
A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
– Daniel Schepler
Nov 15 at 20:01
This is a very good comment. Thanks a lot.
– hamam_Abdallah
Nov 16 at 18:02
This is a very good comment. Thanks a lot.
– hamam_Abdallah
Nov 16 at 18:02
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $$x_n=frac{1}{n}$$
and
$$y_n=frac{1}{n+1}$$
be two sequences of reals in $(0,2)$.
it is clear that
$$lim_{nto+infty}(y_n-x_n)=0$$
but
$$f(y_n)-f(x_n)=(n+1)^5-n^5$$
and
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty (ne 0)$$
We can conclude that $f$ is not uniformly continuous at $(0,2)$.
With $epsilon$.
Take $epsilon=1$ for example.
then for each $eta>0$,
and very very large $n$, we have
$$|y_n-x_n|<eta$$
because
$$lim_{nto+infty}(y_n-x_n)=0$$
and
$$|f(y_n)-f(x_n)|>1$$
because
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty$$
answered Nov 15 at 19:50
hamam_Abdallah
37k21533
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $$x_n=frac{1}{n}$$
and
$$y_n=frac{1}{n+1}$$
be two sequences of reals in $(0,2)$.
it is clear that
$$lim_{nto+infty}(y_n-x_n)=0$$
but
$$f(y_n)-f(x_n)=(n+1)^5-n^5$$
and
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty (ne 0)$$
We can conclude that $f$ is not uniformly continuous at $(0,2)$.
With $epsilon$.
Take $epsilon=1$ for example.
then for each $eta>0$,
and very very large $n$, we have
$$|y_n-x_n|<eta$$
because
$$lim_{nto+infty}(y_n-x_n)=0$$
and
$$|f(y_n)-f(x_n)|>1$$
because
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty$$
answered Nov 15 at 19:50
hamam_Abdallah
37k21533
add a comment |
up vote
2
down vote
accepted
Let $$x_n=frac{1}{n}$$
and
$$y_n=frac{1}{n+1}$$
be two sequences of reals in $(0,2)$.
it is clear that
$$lim_{nto+infty}(y_n-x_n)=0$$
but
$$f(y_n)-f(x_n)=(n+1)^5-n^5$$
and
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty (ne 0)$$
We can conclude that $f$ is not uniformly continuous at $(0,2)$.
With $epsilon$.
Take $epsilon=1$ for example.
then for each $eta>0$,
and very very large $n$, we have
$$|y_n-x_n|<eta$$
because
$$lim_{nto+infty}(y_n-x_n)=0$$
and
$$|f(y_n)-f(x_n)|>1$$
because
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty$$
answered Nov 15 at 19:50
hamam_Abdallah
37k21533
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $$x_n=frac{1}{n}$$
and
$$y_n=frac{1}{n+1}$$
be two sequences of reals in $(0,2)$.
it is clear that
$$lim_{nto+infty}(y_n-x_n)=0$$
but
$$f(y_n)-f(x_n)=(n+1)^5-n^5$$
and
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty (ne 0)$$
We can conclude that $f$ is not uniformly continuous at $(0,2)$.
With $epsilon$.
Take $epsilon=1$ for example.
then for each $eta>0$,
and very very large $n$, we have
$$|y_n-x_n|<eta$$
because
$$lim_{nto+infty}(y_n-x_n)=0$$
and
$$|f(y_n)-f(x_n)|>1$$
because
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty$$
answered Nov 15 at 19:50
hamam_Abdallah
37k21533
Let $$x_n=frac{1}{n}$$
and
$$y_n=frac{1}{n+1}$$
be two sequences of reals in $(0,2)$.
it is clear that
$$lim_{nto+infty}(y_n-x_n)=0$$
but
$$f(y_n)-f(x_n)=(n+1)^5-n^5$$
and
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty (ne 0)$$
We can conclude that $f$ is not uniformly continuous at $(0,2)$.
With $epsilon$.
Take $epsilon=1$ for example.
then for each $eta>0$,
and very very large $n$, we have
$$|y_n-x_n|<eta$$
because
$$lim_{nto+infty}(y_n-x_n)=0$$
and
$$|f(y_n)-f(x_n)|>1$$
because
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty$$
answered Nov 15 at 19:50
hamam_Abdallah
37k21533
edited Nov 16 at 18:01
answered Nov 15 at 19:50
hamam_Abdallah
37k21533
answered Nov 15 at 19:50
hamam_Abdallah
37k21533
answered Nov 15 at 19:50
hamam_Abdallah
37k21533
37k21533
add a comment |
add a comment |
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$(0,2)$...................
– Nosrati
Nov 15 at 19:48
3
A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
– Daniel Schepler
Nov 15 at 20:01
This is a very good comment. Thanks a lot.
– hamam_Abdallah
Nov 16 at 18:02