Is this function uniformly continuous or not?












2














I've been recently introduced to the concept of uniform continuity and I'm having some trouble deciding whether a function is uniformly continuous or not.
The function is:
$$f(x)=begin{cases}dfrac{sin(x^5)}{x} & xneq0 \ 0 & x=0end{cases}$$



I mean, I know that if the derivative of the function is bounded, then we can conclude that it is uniformly continuous, but if it isn't, I really don't know how to proceed.






real-analysis







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  • 2




    Uniformly continuous on what set?
    – user587192
    47 mins ago












  • I believe it must be on $mathbb R$
    – fonini
    46 mins ago










  • On all the real numbers
    – Sergamar
    46 mins ago










  • It can't be on $mathbb{R}$: this function not defined at $x=0$.
    – user587192
    45 mins ago








  • 1




    0 if x = 0, that expresion otherwise
    – Sergamar
    44 mins ago
















2














I've been recently introduced to the concept of uniform continuity and I'm having some trouble deciding whether a function is uniformly continuous or not.
The function is:
$$f(x)=begin{cases}dfrac{sin(x^5)}{x} & xneq0 \ 0 & x=0end{cases}$$



I mean, I know that if the derivative of the function is bounded, then we can conclude that it is uniformly continuous, but if it isn't, I really don't know how to proceed.






real-analysis







share|cite|improve this question









New contributor




Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    Uniformly continuous on what set?
    – user587192
    47 mins ago












  • I believe it must be on $mathbb R$
    – fonini
    46 mins ago










  • On all the real numbers
    – Sergamar
    46 mins ago










  • It can't be on $mathbb{R}$: this function not defined at $x=0$.
    – user587192
    45 mins ago








  • 1




    0 if x = 0, that expresion otherwise
    – Sergamar
    44 mins ago














2












2








2







I've been recently introduced to the concept of uniform continuity and I'm having some trouble deciding whether a function is uniformly continuous or not.
The function is:
$$f(x)=begin{cases}dfrac{sin(x^5)}{x} & xneq0 \ 0 & x=0end{cases}$$



I mean, I know that if the derivative of the function is bounded, then we can conclude that it is uniformly continuous, but if it isn't, I really don't know how to proceed.






real-analysis







share|cite|improve this question









New contributor




Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I've been recently introduced to the concept of uniform continuity and I'm having some trouble deciding whether a function is uniformly continuous or not.
The function is:
$$f(x)=begin{cases}dfrac{sin(x^5)}{x} & xneq0 \ 0 & x=0end{cases}$$



I mean, I know that if the derivative of the function is bounded, then we can conclude that it is uniformly continuous, but if it isn't, I really don't know how to proceed.






real-analysis




real-analysis






share|cite|improve this question









New contributor




Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 40 mins ago





















New contributor




Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 48 mins ago









Sergamar

113




113




New contributor




Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 2




    Uniformly continuous on what set?
    – user587192
    47 mins ago












  • I believe it must be on $mathbb R$
    – fonini
    46 mins ago










  • On all the real numbers
    – Sergamar
    46 mins ago










  • It can't be on $mathbb{R}$: this function not defined at $x=0$.
    – user587192
    45 mins ago








  • 1




    0 if x = 0, that expresion otherwise
    – Sergamar
    44 mins ago














  • 2




    Uniformly continuous on what set?
    – user587192
    47 mins ago












  • I believe it must be on $mathbb R$
    – fonini
    46 mins ago










  • On all the real numbers
    – Sergamar
    46 mins ago










  • It can't be on $mathbb{R}$: this function not defined at $x=0$.
    – user587192
    45 mins ago








  • 1




    0 if x = 0, that expresion otherwise
    – Sergamar
    44 mins ago








2




2




Uniformly continuous on what set?
– user587192
47 mins ago






Uniformly continuous on what set?
– user587192
47 mins ago














I believe it must be on $mathbb R$
– fonini
46 mins ago




I believe it must be on $mathbb R$
– fonini
46 mins ago












On all the real numbers
– Sergamar
46 mins ago




On all the real numbers
– Sergamar
46 mins ago












It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
45 mins ago






It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
45 mins ago






1




1




0 if x = 0, that expresion otherwise
– Sergamar
44 mins ago




0 if x = 0, that expresion otherwise
– Sergamar
44 mins ago










1 Answer
1






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oldest

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3














The function $f$ is uniformly continuous on $mathbb{R}$.



Note that $frac{sin x^5}{x} = x^4 frac{sin x^5}{x^5} to 0 cdot 1 = 0$ as $x to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.



Since $f(x) to 0 $ as $|x| to infty$ the function is also uniformly continuous on intervals $(-infty,-a]$ and $[a,infty)$. Any continuous function with a finite limit as $x to pm infty$ is uniformly continuous -- proved many times on this site.



It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| to +infty$ as $|x| to +infty$ guarantees non-uniform convergence.






share|cite|improve this answer





















  • Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
    – Jakobian
    3 mins ago













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1 Answer
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3














The function $f$ is uniformly continuous on $mathbb{R}$.



Note that $frac{sin x^5}{x} = x^4 frac{sin x^5}{x^5} to 0 cdot 1 = 0$ as $x to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.



Since $f(x) to 0 $ as $|x| to infty$ the function is also uniformly continuous on intervals $(-infty,-a]$ and $[a,infty)$. Any continuous function with a finite limit as $x to pm infty$ is uniformly continuous -- proved many times on this site.



It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| to +infty$ as $|x| to +infty$ guarantees non-uniform convergence.






share|cite|improve this answer





















  • Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
    – Jakobian
    3 mins ago


















3














The function $f$ is uniformly continuous on $mathbb{R}$.



Note that $frac{sin x^5}{x} = x^4 frac{sin x^5}{x^5} to 0 cdot 1 = 0$ as $x to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.



Since $f(x) to 0 $ as $|x| to infty$ the function is also uniformly continuous on intervals $(-infty,-a]$ and $[a,infty)$. Any continuous function with a finite limit as $x to pm infty$ is uniformly continuous -- proved many times on this site.



It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| to +infty$ as $|x| to +infty$ guarantees non-uniform convergence.






share|cite|improve this answer





















  • Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
    – Jakobian
    3 mins ago
















3












3








3






The function $f$ is uniformly continuous on $mathbb{R}$.



Note that $frac{sin x^5}{x} = x^4 frac{sin x^5}{x^5} to 0 cdot 1 = 0$ as $x to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.



Since $f(x) to 0 $ as $|x| to infty$ the function is also uniformly continuous on intervals $(-infty,-a]$ and $[a,infty)$. Any continuous function with a finite limit as $x to pm infty$ is uniformly continuous -- proved many times on this site.



It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| to +infty$ as $|x| to +infty$ guarantees non-uniform convergence.






share|cite|improve this answer












The function $f$ is uniformly continuous on $mathbb{R}$.



Note that $frac{sin x^5}{x} = x^4 frac{sin x^5}{x^5} to 0 cdot 1 = 0$ as $x to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.



Since $f(x) to 0 $ as $|x| to infty$ the function is also uniformly continuous on intervals $(-infty,-a]$ and $[a,infty)$. Any continuous function with a finite limit as $x to pm infty$ is uniformly continuous -- proved many times on this site.



It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| to +infty$ as $|x| to +infty$ guarantees non-uniform convergence.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 12 mins ago









RRL

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48.7k42573












  • Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
    – Jakobian
    3 mins ago




















  • Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
    – Jakobian
    3 mins ago


















Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
– Jakobian
3 mins ago






Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
– Jakobian
3 mins ago












Sergamar is a new contributor. Be nice, and check out our Code of Conduct.










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