Determine if the function is uniformly continuous
$begingroup$
Determine if the function: $$f(x)=frac{x}{(1-x)^2}$$ $forall x>1$ is uniformly continuous.
I know that it is not. But I'm having trouble proving it.
I reach to a point that i get this: $$frac{|x-y||1-xy|}{(1-x)^2(1-y)^2} le delta|1-xy|$$ and I cannot continue.
Any help would be appreciated!
continuity uniform-continuity
edited Dec 18 '18 at 14:05
idea
2,16841125
asked Dec 18 '18 at 14:00
argiriskarargiriskar
1409
$endgroup$
add a comment |
$begingroup$
Determine if the function: $$f(x)=frac{x}{(1-x)^2}$$ $forall x>1$ is uniformly continuous.
I know that it is not. But I'm having trouble proving it.
I reach to a point that i get this: $$frac{|x-y||1-xy|}{(1-x)^2(1-y)^2} le delta|1-xy|$$ and I cannot continue.
Any help would be appreciated!
continuity uniform-continuity
edited Dec 18 '18 at 14:05
idea
2,16841125
asked Dec 18 '18 at 14:00
argiriskarargiriskar
1409
$endgroup$
1
$begingroup$
Try finding a Cauchy sequence $x_n$ that fails to map to a Cauchy sequence. Try thinking of a sequence $x_n$ that converges to $1$, but is strictly greater than $1$.
$endgroup$
– Theo Bendit
Dec 18 '18 at 14:06
$begingroup$
wolframalpha.com/input/?i=x%2F(1-x)%5E2 may help
$endgroup$
– idea
Dec 18 '18 at 14:08
add a comment |
$begingroup$
Determine if the function: $$f(x)=frac{x}{(1-x)^2}$$ $forall x>1$ is uniformly continuous.
I know that it is not. But I'm having trouble proving it.
I reach to a point that i get this: $$frac{|x-y||1-xy|}{(1-x)^2(1-y)^2} le delta|1-xy|$$ and I cannot continue.
Any help would be appreciated!
continuity uniform-continuity
edited Dec 18 '18 at 14:05
idea
2,16841125
asked Dec 18 '18 at 14:00
argiriskarargiriskar
1409
$endgroup$
Determine if the function: $$f(x)=frac{x}{(1-x)^2}$$ $forall x>1$ is uniformly continuous.
I know that it is not. But I'm having trouble proving it.
I reach to a point that i get this: $$frac{|x-y||1-xy|}{(1-x)^2(1-y)^2} le delta|1-xy|$$ and I cannot continue.
Any help would be appreciated!
continuity uniform-continuity
continuity uniform-continuity
edited Dec 18 '18 at 14:05
idea
2,16841125
asked Dec 18 '18 at 14:00
argiriskarargiriskar
1409
edited Dec 18 '18 at 14:05
idea
2,16841125
asked Dec 18 '18 at 14:00
argiriskarargiriskar
1409
edited Dec 18 '18 at 14:05
idea
2,16841125
edited Dec 18 '18 at 14:05
idea
2,16841125
edited Dec 18 '18 at 14:05
idea
2,16841125
2,16841125
asked Dec 18 '18 at 14:00
argiriskarargiriskar
1409
asked Dec 18 '18 at 14:00
argiriskarargiriskar
1409
asked Dec 18 '18 at 14:00
argiriskarargiriskar
1409
1409
1
$begingroup$
Try finding a Cauchy sequence $x_n$ that fails to map to a Cauchy sequence. Try thinking of a sequence $x_n$ that converges to $1$, but is strictly greater than $1$.
$endgroup$
– Theo Bendit
Dec 18 '18 at 14:06
$begingroup$
wolframalpha.com/input/?i=x%2F(1-x)%5E2 may help
$endgroup$
– idea
Dec 18 '18 at 14:08
add a comment |
1
$begingroup$
Try finding a Cauchy sequence $x_n$ that fails to map to a Cauchy sequence. Try thinking of a sequence $x_n$ that converges to $1$, but is strictly greater than $1$.
$endgroup$
– Theo Bendit
Dec 18 '18 at 14:06
$begingroup$
wolframalpha.com/input/?i=x%2F(1-x)%5E2 may help
$endgroup$
– idea
Dec 18 '18 at 14:08
1
1
$begingroup$
Try finding a Cauchy sequence $x_n$ that fails to map to a Cauchy sequence. Try thinking of a sequence $x_n$ that converges to $1$, but is strictly greater than $1$.
$endgroup$
– Theo Bendit
Dec 18 '18 at 14:06
$begingroup$
Try finding a Cauchy sequence $x_n$ that fails to map to a Cauchy sequence. Try thinking of a sequence $x_n$ that converges to $1$, but is strictly greater than $1$.
$endgroup$
– Theo Bendit
Dec 18 '18 at 14:06
$begingroup$
wolframalpha.com/input/?i=x%2F(1-x)%5E2 may help
$endgroup$
– idea
Dec 18 '18 at 14:08
$begingroup$
wolframalpha.com/input/?i=x%2F(1-x)%5E2 may help
$endgroup$
– idea
Dec 18 '18 at 14:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You know that $lim_{xto1^+}f(x)=infty$. So, for any $delta>0$, you can find $x,yin(1,1+delta)$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$. But $lvert x-yrvert<delta$.
answered Dec 18 '18 at 14:06
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
Do we need to prove that in a neighborhood of 1 we get $|f(x)-f(y)|ge 1$?
$endgroup$
– argiriskar
Dec 18 '18 at 14:11
$begingroup$
I didn't say that. Actually, I don't even know what it means. What I wrote was that in $(1,1+delta)$ there are elements $x$ and $y$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$.
$endgroup$
– José Carlos Santos
Dec 18 '18 at 14:15
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You know that $lim_{xto1^+}f(x)=infty$. So, for any $delta>0$, you can find $x,yin(1,1+delta)$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$. But $lvert x-yrvert<delta$.
answered Dec 18 '18 at 14:06
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
Do we need to prove that in a neighborhood of 1 we get $|f(x)-f(y)|ge 1$?
$endgroup$
– argiriskar
Dec 18 '18 at 14:11
$begingroup$
I didn't say that. Actually, I don't even know what it means. What I wrote was that in $(1,1+delta)$ there are elements $x$ and $y$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$.
$endgroup$
– José Carlos Santos
Dec 18 '18 at 14:15
add a comment |
$begingroup$
You know that $lim_{xto1^+}f(x)=infty$. So, for any $delta>0$, you can find $x,yin(1,1+delta)$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$. But $lvert x-yrvert<delta$.
answered Dec 18 '18 at 14:06
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
Do we need to prove that in a neighborhood of 1 we get $|f(x)-f(y)|ge 1$?
$endgroup$
– argiriskar
Dec 18 '18 at 14:11
$begingroup$
I didn't say that. Actually, I don't even know what it means. What I wrote was that in $(1,1+delta)$ there are elements $x$ and $y$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$.
$endgroup$
– José Carlos Santos
Dec 18 '18 at 14:15
add a comment |
$begingroup$
You know that $lim_{xto1^+}f(x)=infty$. So, for any $delta>0$, you can find $x,yin(1,1+delta)$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$. But $lvert x-yrvert<delta$.
answered Dec 18 '18 at 14:06
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
You know that $lim_{xto1^+}f(x)=infty$. So, for any $delta>0$, you can find $x,yin(1,1+delta)$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$. But $lvert x-yrvert<delta$.
answered Dec 18 '18 at 14:06
José Carlos SantosJosé Carlos Santos
163k22130233
answered Dec 18 '18 at 14:06
José Carlos SantosJosé Carlos Santos
163k22130233
answered Dec 18 '18 at 14:06
José Carlos SantosJosé Carlos Santos
163k22130233
answered Dec 18 '18 at 14:06
José Carlos SantosJosé Carlos Santos
163k22130233
163k22130233
$begingroup$
Do we need to prove that in a neighborhood of 1 we get $|f(x)-f(y)|ge 1$?
$endgroup$
– argiriskar
Dec 18 '18 at 14:11
$begingroup$
I didn't say that. Actually, I don't even know what it means. What I wrote was that in $(1,1+delta)$ there are elements $x$ and $y$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$.
$endgroup$
– José Carlos Santos
Dec 18 '18 at 14:15
add a comment |
$begingroup$
Do we need to prove that in a neighborhood of 1 we get $|f(x)-f(y)|ge 1$?
$endgroup$
– argiriskar
Dec 18 '18 at 14:11
$begingroup$
I didn't say that. Actually, I don't even know what it means. What I wrote was that in $(1,1+delta)$ there are elements $x$ and $y$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$.
$endgroup$
– José Carlos Santos
Dec 18 '18 at 14:15
$begingroup$
Do we need to prove that in a neighborhood of 1 we get $|f(x)-f(y)|ge 1$?
$endgroup$
– argiriskar
Dec 18 '18 at 14:11
$begingroup$
Do we need to prove that in a neighborhood of 1 we get $|f(x)-f(y)|ge 1$?
$endgroup$
– argiriskar
Dec 18 '18 at 14:11
$begingroup$
I didn't say that. Actually, I don't even know what it means. What I wrote was that in $(1,1+delta)$ there are elements $x$ and $y$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$.
$endgroup$
– José Carlos Santos
Dec 18 '18 at 14:15
$begingroup$
I didn't say that. Actually, I don't even know what it means. What I wrote was that in $(1,1+delta)$ there are elements $x$ and $y$ such that $bigllvert f(x)-f(y)bigrrvertgeqslant1$.
$endgroup$
– José Carlos Santos
Dec 18 '18 at 14:15
add a comment |
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$begingroup$
Try finding a Cauchy sequence $x_n$ that fails to map to a Cauchy sequence. Try thinking of a sequence $x_n$ that converges to $1$, but is strictly greater than $1$.
$endgroup$
– Theo Bendit
Dec 18 '18 at 14:06
$begingroup$
wolframalpha.com/input/?i=x%2F(1-x)%5E2 may help
$endgroup$
– idea
Dec 18 '18 at 14:08