Examine the uniform continuity $f(x)=(e^x-1)sinfrac13^x-1$ for $x in (0,+infty)$ The 2019 Stack Overflow Developer Survey Results Are InConvergence of Integral Implies Uniform convergence of Equicontinuous FamilyUniform Continuity: i'm right?Study the uniform continuity of these functions:If a non-decreasing function $f: mathbbRrightarrow (0,+infty)$ satisfies $liminf (f(n+1)-f(n))>0$, then $lim sup fracf(x)x>0$Proving uniform continuity and uniform discontinuityhelp showing the uniform continuity function FHaving troubles in understanding subsequential limits: what's $displaystylelim_nrightarrowinfty sup5+frac1n sin(n)$?Uniform Continuity for $sin(frac1x+1)$Uniform Continuity of function $f(x)=sqrt xsinsqrt x$ on $[0,infty)$Investigate whether $ f $ meets Lipschitz continuity and whether it is uniform continuity
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Examine the uniform continuity $f(x)=(e^x-1)sinfrac13^x-1$ for $x in (0,+infty)$
The 2019 Stack Overflow Developer Survey Results Are InConvergence of Integral Implies Uniform convergence of Equicontinuous FamilyUniform Continuity: i'm right?Study the uniform continuity of these functions:If a non-decreasing function $f: mathbbRrightarrow (0,+infty)$ satisfies $liminf (f(n+1)-f(n))>0$, then $lim sup fracf(x)x>0$Proving uniform continuity and uniform discontinuityhelp showing the uniform continuity function FHaving troubles in understanding subsequential limits: what's $displaystylelim_nrightarrowinfty sup5+frac1n sin(n)$?Uniform Continuity for $sin(frac1x+1)$Uniform Continuity of function $f(x)=sqrt xsinsqrt x$ on $[0,infty)$Investigate whether $ f $ meets Lipschitz continuity and whether it is uniform continuity
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My try:
I know that if for all $x_n,y_n$ condidion $lim_n rightarrow +infty(x_n-y_n)=0$ implies $lim_n rightarrow +infty(f(x_n)-f(y_n)=0$ then $f$ is uniform continuity.
Let: $$x_n=frac1n$$ $$y_n=frac12n$$Then:$$x_n-y_n=frac12n rightarrow 0$$
$$f(x_n)-f(y_n)=(e^frac1n-1)sinfrac13^frac1n-1-(e^frac12n-1)sinfrac13^frac12n-1$$
I know that $f(x_n)-f(y_n)$ is periodic so $lim neq 0$ so I find counterexample that $f$ is uniform continuity.
However I don't know how I can show it in a professional way.
Have you some idea?
real-analysis
asked Mar 30 at 17:04
MP3129MP3129
787211
$endgroup$
add a comment |
$begingroup$
My try:
I know that if for all $x_n,y_n$ condidion $lim_n rightarrow +infty(x_n-y_n)=0$ implies $lim_n rightarrow +infty(f(x_n)-f(y_n)=0$ then $f$ is uniform continuity.
Let: $$x_n=frac1n$$ $$y_n=frac12n$$Then:$$x_n-y_n=frac12n rightarrow 0$$
$$f(x_n)-f(y_n)=(e^frac1n-1)sinfrac13^frac1n-1-(e^frac12n-1)sinfrac13^frac12n-1$$
I know that $f(x_n)-f(y_n)$ is periodic so $lim neq 0$ so I find counterexample that $f$ is uniform continuity.
However I don't know how I can show it in a professional way.
Have you some idea?
real-analysis
asked Mar 30 at 17:04
MP3129MP3129
787211
$endgroup$
$begingroup$
You are just showing for one particular sequence, that is not a proof.
$endgroup$
– copper.hat
Mar 30 at 17:29
$begingroup$
@copper.hat But the OP is trying to show UC fails.
$endgroup$
– zhw.
Mar 30 at 17:30
$begingroup$
@zhw.: It is uniformly continuous.
$endgroup$
– copper.hat
Mar 30 at 17:33
1
$begingroup$
@copper.hat Yes I know, but the OP believes otherwise. So looking for one sequence is decent strategy in attempting to show it's not UC,
$endgroup$
– zhw.
Mar 30 at 17:36
add a comment |
$begingroup$
My try:
I know that if for all $x_n,y_n$ condidion $lim_n rightarrow +infty(x_n-y_n)=0$ implies $lim_n rightarrow +infty(f(x_n)-f(y_n)=0$ then $f$ is uniform continuity.
Let: $$x_n=frac1n$$ $$y_n=frac12n$$Then:$$x_n-y_n=frac12n rightarrow 0$$
$$f(x_n)-f(y_n)=(e^frac1n-1)sinfrac13^frac1n-1-(e^frac12n-1)sinfrac13^frac12n-1$$
I know that $f(x_n)-f(y_n)$ is periodic so $lim neq 0$ so I find counterexample that $f$ is uniform continuity.
However I don't know how I can show it in a professional way.
Have you some idea?
real-analysis
asked Mar 30 at 17:04
MP3129MP3129
787211
$endgroup$
My try:
I know that if for all $x_n,y_n$ condidion $lim_n rightarrow +infty(x_n-y_n)=0$ implies $lim_n rightarrow +infty(f(x_n)-f(y_n)=0$ then $f$ is uniform continuity.
Let: $$x_n=frac1n$$ $$y_n=frac12n$$Then:$$x_n-y_n=frac12n rightarrow 0$$
$$f(x_n)-f(y_n)=(e^frac1n-1)sinfrac13^frac1n-1-(e^frac12n-1)sinfrac13^frac12n-1$$
I know that $f(x_n)-f(y_n)$ is periodic so $lim neq 0$ so I find counterexample that $f$ is uniform continuity.
However I don't know how I can show it in a professional way.
Have you some idea?
real-analysis
real-analysis
asked Mar 30 at 17:04
MP3129MP3129
787211
asked Mar 30 at 17:04
MP3129MP3129
787211
asked Mar 30 at 17:04
MP3129MP3129
787211
asked Mar 30 at 17:04
MP3129MP3129
787211
asked Mar 30 at 17:04
MP3129MP3129
787211
787211
$begingroup$
You are just showing for one particular sequence, that is not a proof.
$endgroup$
– copper.hat
Mar 30 at 17:29
$begingroup$
@copper.hat But the OP is trying to show UC fails.
$endgroup$
– zhw.
Mar 30 at 17:30
$begingroup$
@zhw.: It is uniformly continuous.
$endgroup$
– copper.hat
Mar 30 at 17:33
1
$begingroup$
@copper.hat Yes I know, but the OP believes otherwise. So looking for one sequence is decent strategy in attempting to show it's not UC,
$endgroup$
– zhw.
Mar 30 at 17:36
add a comment |
$begingroup$
You are just showing for one particular sequence, that is not a proof.
$endgroup$
– copper.hat
Mar 30 at 17:29
$begingroup$
@copper.hat But the OP is trying to show UC fails.
$endgroup$
– zhw.
Mar 30 at 17:30
$begingroup$
@zhw.: It is uniformly continuous.
$endgroup$
– copper.hat
Mar 30 at 17:33
1
$begingroup$
@copper.hat Yes I know, but the OP believes otherwise. So looking for one sequence is decent strategy in attempting to show it's not UC,
$endgroup$
– zhw.
Mar 30 at 17:36
$begingroup$
You are just showing for one particular sequence, that is not a proof.
$endgroup$
– copper.hat
Mar 30 at 17:29
$begingroup$
You are just showing for one particular sequence, that is not a proof.
$endgroup$
– copper.hat
Mar 30 at 17:29
$begingroup$
@copper.hat But the OP is trying to show UC fails.
$endgroup$
– zhw.
Mar 30 at 17:30
$begingroup$
@copper.hat But the OP is trying to show UC fails.
$endgroup$
– zhw.
Mar 30 at 17:30
$begingroup$
@zhw.: It is uniformly continuous.
$endgroup$
– copper.hat
Mar 30 at 17:33
$begingroup$
@zhw.: It is uniformly continuous.
$endgroup$
– copper.hat
Mar 30 at 17:33
1
1
$begingroup$
@copper.hat Yes I know, but the OP believes otherwise. So looking for one sequence is decent strategy in attempting to show it's not UC,
$endgroup$
– zhw.
Mar 30 at 17:36
$begingroup$
@copper.hat Yes I know, but the OP believes otherwise. So looking for one sequence is decent strategy in attempting to show it's not UC,
$endgroup$
– zhw.
Mar 30 at 17:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Show that $lim_x downarrow 0 f(x) = 0$, $lim_x to infty f(x) = 0$ and combine
this with uniform continuity of continuous functions on compact intervals to show that
$f$ is uniformly continuous.
answered Mar 30 at 17:38
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
This limit is equal to zero because
$$lim_ntoinfty (e^frac1kn-1)=0$$
given that $kinmathbbR$, $kne0$. As $sin(x)$ is bounded for any $xinmathbbR$ the value of the two multiplied together is still zero. Also $f(x_n)-f(y_n)$ is not periodic.
answered Mar 30 at 17:17
Peter ForemanPeter Foreman
6,9401318
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Show that $lim_x downarrow 0 f(x) = 0$, $lim_x to infty f(x) = 0$ and combine
this with uniform continuity of continuous functions on compact intervals to show that
$f$ is uniformly continuous.
answered Mar 30 at 17:38
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
Show that $lim_x downarrow 0 f(x) = 0$, $lim_x to infty f(x) = 0$ and combine
this with uniform continuity of continuous functions on compact intervals to show that
$f$ is uniformly continuous.
answered Mar 30 at 17:38
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
Show that $lim_x downarrow 0 f(x) = 0$, $lim_x to infty f(x) = 0$ and combine
this with uniform continuity of continuous functions on compact intervals to show that
$f$ is uniformly continuous.
answered Mar 30 at 17:38
copper.hatcopper.hat
128k561161
$endgroup$
Show that $lim_x downarrow 0 f(x) = 0$, $lim_x to infty f(x) = 0$ and combine
this with uniform continuity of continuous functions on compact intervals to show that
$f$ is uniformly continuous.
answered Mar 30 at 17:38
copper.hatcopper.hat
128k561161
answered Mar 30 at 17:38
copper.hatcopper.hat
128k561161
answered Mar 30 at 17:38
copper.hatcopper.hat
128k561161
answered Mar 30 at 17:38
copper.hatcopper.hat
128k561161
128k561161
add a comment |
add a comment |
$begingroup$
This limit is equal to zero because
$$lim_ntoinfty (e^frac1kn-1)=0$$
given that $kinmathbbR$, $kne0$. As $sin(x)$ is bounded for any $xinmathbbR$ the value of the two multiplied together is still zero. Also $f(x_n)-f(y_n)$ is not periodic.
answered Mar 30 at 17:17
Peter ForemanPeter Foreman
6,9401318
$endgroup$
add a comment |
$begingroup$
This limit is equal to zero because
$$lim_ntoinfty (e^frac1kn-1)=0$$
given that $kinmathbbR$, $kne0$. As $sin(x)$ is bounded for any $xinmathbbR$ the value of the two multiplied together is still zero. Also $f(x_n)-f(y_n)$ is not periodic.
answered Mar 30 at 17:17
Peter ForemanPeter Foreman
6,9401318
$endgroup$
add a comment |
$begingroup$
This limit is equal to zero because
$$lim_ntoinfty (e^frac1kn-1)=0$$
given that $kinmathbbR$, $kne0$. As $sin(x)$ is bounded for any $xinmathbbR$ the value of the two multiplied together is still zero. Also $f(x_n)-f(y_n)$ is not periodic.
answered Mar 30 at 17:17
Peter ForemanPeter Foreman
6,9401318
$endgroup$
This limit is equal to zero because
$$lim_ntoinfty (e^frac1kn-1)=0$$
given that $kinmathbbR$, $kne0$. As $sin(x)$ is bounded for any $xinmathbbR$ the value of the two multiplied together is still zero. Also $f(x_n)-f(y_n)$ is not periodic.
answered Mar 30 at 17:17
Peter ForemanPeter Foreman
6,9401318
answered Mar 30 at 17:17
Peter ForemanPeter Foreman
6,9401318
answered Mar 30 at 17:17
Peter ForemanPeter Foreman
6,9401318
answered Mar 30 at 17:17
Peter ForemanPeter Foreman
6,9401318
6,9401318
add a comment |
add a comment |
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$begingroup$
You are just showing for one particular sequence, that is not a proof.
$endgroup$
– copper.hat
Mar 30 at 17:29
$begingroup$
@copper.hat But the OP is trying to show UC fails.
$endgroup$
– zhw.
Mar 30 at 17:30
$begingroup$
@zhw.: It is uniformly continuous.
$endgroup$
– copper.hat
Mar 30 at 17:33
1
$begingroup$
@copper.hat Yes I know, but the OP believes otherwise. So looking for one sequence is decent strategy in attempting to show it's not UC,
$endgroup$
– zhw.
Mar 30 at 17:36