Analyse Equicontinuity (uniform continuity) of a function $ frac{sin(x)}{x},; x > 0 $












0












$begingroup$


How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$



When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?






calculus continuity equicontinuity







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  • 1




    $begingroup$
    By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:39










  • $begingroup$
    I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
    $endgroup$
    – R0xx0rZzz
    Dec 9 '18 at 17:41












  • $begingroup$
    Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:55
















0












$begingroup$


How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$



When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?






calculus continuity equicontinuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:39










  • $begingroup$
    I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
    $endgroup$
    – R0xx0rZzz
    Dec 9 '18 at 17:41












  • $begingroup$
    Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:55














0












0








0





$begingroup$


How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$



When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?






calculus continuity equicontinuity







share|cite|improve this question











$endgroup$




How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$



When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?






calculus continuity equicontinuity




calculus continuity equicontinuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 23:37









user376343

3,3833826




3,3833826










asked Dec 9 '18 at 17:31









R0xx0rZzzR0xx0rZzz

536




536








  • 1




    $begingroup$
    By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:39










  • $begingroup$
    I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
    $endgroup$
    – R0xx0rZzz
    Dec 9 '18 at 17:41












  • $begingroup$
    Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:55














  • 1




    $begingroup$
    By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:39










  • $begingroup$
    I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
    $endgroup$
    – R0xx0rZzz
    Dec 9 '18 at 17:41












  • $begingroup$
    Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:55








1




1




$begingroup$
By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:39




$begingroup$
By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:39












$begingroup$
I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
$endgroup$
– R0xx0rZzz
Dec 9 '18 at 17:41






$begingroup$
I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
$endgroup$
– R0xx0rZzz
Dec 9 '18 at 17:41














$begingroup$
Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:55




$begingroup$
Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:55










2 Answers
2






active

oldest

votes


















1












$begingroup$

This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.



Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 19:07





















1












$begingroup$

If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.






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  • $begingroup$
    typo corrected$,$
    $endgroup$
    – zhw.
    Dec 24 '18 at 17:06











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.



Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 19:07


















1












$begingroup$

This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.



Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 19:07
















1












1








1





$begingroup$

This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.



Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.






share|cite|improve this answer









$endgroup$



This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.



Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 18:32









Christian BlatterChristian Blatter

173k7113326




173k7113326








  • 1




    $begingroup$
    To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 19:07
















  • 1




    $begingroup$
    To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 19:07










1




1




$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07






$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07













1












$begingroup$

If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    typo corrected$,$
    $endgroup$
    – zhw.
    Dec 24 '18 at 17:06
















1












$begingroup$

If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    typo corrected$,$
    $endgroup$
    – zhw.
    Dec 24 '18 at 17:06














1












1








1





$begingroup$

If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.






share|cite|improve this answer











$endgroup$



If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 17:05

























answered Dec 10 '18 at 0:10









zhw.zhw.

72.1k43075




72.1k43075












  • $begingroup$
    typo corrected$,$
    $endgroup$
    – zhw.
    Dec 24 '18 at 17:06


















  • $begingroup$
    typo corrected$,$
    $endgroup$
    – zhw.
    Dec 24 '18 at 17:06
















$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06




$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06


















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