When teaching someone how to prove a function is uniformly continuous, using epsilon/delta, which example...












1












$begingroup$


I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is continuous over an open interval.



Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.



Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?






calculus limits







share|improve this question









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  • 2




    $begingroup$
    A linear function, perhaps?
    $endgroup$
    – paw88789
    4 hours ago










  • $begingroup$
    @paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
    $endgroup$
    – Alec
    4 hours ago










  • $begingroup$
    $|sin x - sin y| le |x-y|$ makes sine a good candidate.
    $endgroup$
    – user3813
    49 secs ago
















1












$begingroup$


I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is continuous over an open interval.



Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.



Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?






calculus limits







share|improve this question









$endgroup$








  • 2




    $begingroup$
    A linear function, perhaps?
    $endgroup$
    – paw88789
    4 hours ago










  • $begingroup$
    @paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
    $endgroup$
    – Alec
    4 hours ago










  • $begingroup$
    $|sin x - sin y| le |x-y|$ makes sine a good candidate.
    $endgroup$
    – user3813
    49 secs ago














1












1








1





$begingroup$


I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is continuous over an open interval.



Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.



Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?






calculus limits







share|improve this question









$endgroup$




I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is continuous over an open interval.



Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.



Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?






calculus limits




calculus limits






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 4 hours ago









AlecAlec

609310




609310








  • 2




    $begingroup$
    A linear function, perhaps?
    $endgroup$
    – paw88789
    4 hours ago










  • $begingroup$
    @paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
    $endgroup$
    – Alec
    4 hours ago










  • $begingroup$
    $|sin x - sin y| le |x-y|$ makes sine a good candidate.
    $endgroup$
    – user3813
    49 secs ago














  • 2




    $begingroup$
    A linear function, perhaps?
    $endgroup$
    – paw88789
    4 hours ago










  • $begingroup$
    @paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
    $endgroup$
    – Alec
    4 hours ago










  • $begingroup$
    $|sin x - sin y| le |x-y|$ makes sine a good candidate.
    $endgroup$
    – user3813
    49 secs ago








2




2




$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
4 hours ago




$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
4 hours ago












$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
4 hours ago




$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
4 hours ago












$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
49 secs ago




$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
49 secs ago










1 Answer
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$begingroup$

I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.






share|improve this answer









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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    2












    $begingroup$

    I think this cannot be understood without a contrasting example where it fails.
    So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
    It is continuous over that interval, but not uniformly continuous.
    Fix an $epsilon > 0$; then for any $delta > 0$ one can
    arrange the difference in $f$-values to exceed $epsilon$ by getting
    close enough to $x=0$.






    share|improve this answer









    $endgroup$


















      2












      $begingroup$

      I think this cannot be understood without a contrasting example where it fails.
      So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
      It is continuous over that interval, but not uniformly continuous.
      Fix an $epsilon > 0$; then for any $delta > 0$ one can
      arrange the difference in $f$-values to exceed $epsilon$ by getting
      close enough to $x=0$.






      share|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I think this cannot be understood without a contrasting example where it fails.
        So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
        It is continuous over that interval, but not uniformly continuous.
        Fix an $epsilon > 0$; then for any $delta > 0$ one can
        arrange the difference in $f$-values to exceed $epsilon$ by getting
        close enough to $x=0$.






        share|improve this answer









        $endgroup$



        I think this cannot be understood without a contrasting example where it fails.
        So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
        It is continuous over that interval, but not uniformly continuous.
        Fix an $epsilon > 0$; then for any $delta > 0$ one can
        arrange the difference in $f$-values to exceed $epsilon$ by getting
        close enough to $x=0$.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 hours ago









        Joseph O'RourkeJoseph O'Rourke

        15k33280




        15k33280






























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