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.










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  • 2




    Uniformly continuous on what set?
    – user587192
    3 hours ago












  • I believe it must be on $mathbb R$
    – fonini
    3 hours ago










  • On all the real numbers
    – Sergamar
    3 hours ago










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








  • 1




    0 if x = 0, that expresion otherwise
    – Sergamar
    3 hours 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.










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
    3 hours ago












  • I believe it must be on $mathbb R$
    – fonini
    3 hours ago










  • On all the real numbers
    – Sergamar
    3 hours ago










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








  • 1




    0 if x = 0, that expresion otherwise
    – Sergamar
    3 hours 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.










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






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 3 hours ago





















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Check out our Code of Conduct.









asked 3 hours 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
    3 hours ago












  • I believe it must be on $mathbb R$
    – fonini
    3 hours ago










  • On all the real numbers
    – Sergamar
    3 hours ago










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








  • 1




    0 if x = 0, that expresion otherwise
    – Sergamar
    3 hours ago














  • 2




    Uniformly continuous on what set?
    – user587192
    3 hours ago












  • I believe it must be on $mathbb R$
    – fonini
    3 hours ago










  • On all the real numbers
    – Sergamar
    3 hours ago










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








  • 1




    0 if x = 0, that expresion otherwise
    – Sergamar
    3 hours ago








2




2




Uniformly continuous on what set?
– user587192
3 hours ago






Uniformly continuous on what set?
– user587192
3 hours ago














I believe it must be on $mathbb R$
– fonini
3 hours ago




I believe it must be on $mathbb R$
– fonini
3 hours ago












On all the real numbers
– Sergamar
3 hours ago




On all the real numbers
– Sergamar
3 hours ago












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






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






1




1




0 if x = 0, that expresion otherwise
– Sergamar
3 hours ago




0 if x = 0, that expresion otherwise
– Sergamar
3 hours ago










1 Answer
1






active

oldest

votes


















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 continuity.






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
    2 hours ago








  • 2




    @Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
    – RRL
    2 hours ago












  • @Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
    – ImNotTheGuy
    2 hours ago










  • @RRL oh, thank you, it was a typo and it got me confused
    – Jakobian
    2 hours ago












  • @Jakobian It got me confused too, lol.
    – ImNotTheGuy
    2 hours 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 continuity.






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
    2 hours ago








  • 2




    @Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
    – RRL
    2 hours ago












  • @Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
    – ImNotTheGuy
    2 hours ago










  • @RRL oh, thank you, it was a typo and it got me confused
    – Jakobian
    2 hours ago












  • @Jakobian It got me confused too, lol.
    – ImNotTheGuy
    2 hours 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 continuity.






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
    2 hours ago








  • 2




    @Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
    – RRL
    2 hours ago












  • @Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
    – ImNotTheGuy
    2 hours ago










  • @RRL oh, thank you, it was a typo and it got me confused
    – Jakobian
    2 hours ago












  • @Jakobian It got me confused too, lol.
    – ImNotTheGuy
    2 hours 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 continuity.






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 continuity.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 2 hours ago









RRL

48.7k42573




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
    2 hours ago








  • 2




    @Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
    – RRL
    2 hours ago












  • @Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
    – ImNotTheGuy
    2 hours ago










  • @RRL oh, thank you, it was a typo and it got me confused
    – Jakobian
    2 hours ago












  • @Jakobian It got me confused too, lol.
    – ImNotTheGuy
    2 hours 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
    2 hours ago








  • 2




    @Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
    – RRL
    2 hours ago












  • @Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
    – ImNotTheGuy
    2 hours ago










  • @RRL oh, thank you, it was a typo and it got me confused
    – Jakobian
    2 hours ago












  • @Jakobian It got me confused too, lol.
    – ImNotTheGuy
    2 hours 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
2 hours 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
2 hours ago






2




2




@Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
– RRL
2 hours ago






@Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
– RRL
2 hours ago














@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
2 hours ago




@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
2 hours ago












@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
2 hours ago






@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
2 hours ago














@Jakobian It got me confused too, lol.
– ImNotTheGuy
2 hours ago




@Jakobian It got me confused too, lol.
– ImNotTheGuy
2 hours ago










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










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