Show as $xrightarrow 0+, cos(1/x) / x < infty$ [closed]












-1












$begingroup$


I am trying to show that given
$$f(x) = xsin(1/x),quad x>0,quad f(0)=0, $$
the derivative is bounded. Namely, $ f'(x) < infty$
for $ x in (0,a)$ for some $ainmathbb{R}$.



My attempt at a solution lead me to taking the derivative using the Product Rule but I can't find a way to justify that $-frac{1}{x}cos(1/x) < infty$
as $x rightarrow 0^+$ despite Wolfram-Alpha telling me that the limit is indeed $< infty$.






real-analysis limits measure-theory







share|cite|improve this question











$endgroup$



closed as unclear what you're asking by Did, RRL, José Carlos Santos, Leucippus, Cesareo Dec 3 '18 at 10:02


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
    $endgroup$
    – RRL
    Dec 3 '18 at 1:33










  • $begingroup$
    Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
    $endgroup$
    – raka
    Dec 3 '18 at 1:35






  • 3




    $begingroup$
    It is not bounded near $0$
    $endgroup$
    – RRL
    Dec 3 '18 at 1:37










  • $begingroup$
    I see it now. Thank you.
    $endgroup$
    – raka
    Dec 3 '18 at 1:38
















-1












$begingroup$


I am trying to show that given
$$f(x) = xsin(1/x),quad x>0,quad f(0)=0, $$
the derivative is bounded. Namely, $ f'(x) < infty$
for $ x in (0,a)$ for some $ainmathbb{R}$.



My attempt at a solution lead me to taking the derivative using the Product Rule but I can't find a way to justify that $-frac{1}{x}cos(1/x) < infty$
as $x rightarrow 0^+$ despite Wolfram-Alpha telling me that the limit is indeed $< infty$.






real-analysis limits measure-theory







share|cite|improve this question











$endgroup$



closed as unclear what you're asking by Did, RRL, José Carlos Santos, Leucippus, Cesareo Dec 3 '18 at 10:02


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
    $endgroup$
    – RRL
    Dec 3 '18 at 1:33










  • $begingroup$
    Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
    $endgroup$
    – raka
    Dec 3 '18 at 1:35






  • 3




    $begingroup$
    It is not bounded near $0$
    $endgroup$
    – RRL
    Dec 3 '18 at 1:37










  • $begingroup$
    I see it now. Thank you.
    $endgroup$
    – raka
    Dec 3 '18 at 1:38














-1












-1








-1





$begingroup$


I am trying to show that given
$$f(x) = xsin(1/x),quad x>0,quad f(0)=0, $$
the derivative is bounded. Namely, $ f'(x) < infty$
for $ x in (0,a)$ for some $ainmathbb{R}$.



My attempt at a solution lead me to taking the derivative using the Product Rule but I can't find a way to justify that $-frac{1}{x}cos(1/x) < infty$
as $x rightarrow 0^+$ despite Wolfram-Alpha telling me that the limit is indeed $< infty$.






real-analysis limits measure-theory







share|cite|improve this question











$endgroup$




I am trying to show that given
$$f(x) = xsin(1/x),quad x>0,quad f(0)=0, $$
the derivative is bounded. Namely, $ f'(x) < infty$
for $ x in (0,a)$ for some $ainmathbb{R}$.



My attempt at a solution lead me to taking the derivative using the Product Rule but I can't find a way to justify that $-frac{1}{x}cos(1/x) < infty$
as $x rightarrow 0^+$ despite Wolfram-Alpha telling me that the limit is indeed $< infty$.






real-analysis limits measure-theory




real-analysis limits measure-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 1:35







raka

















asked Dec 3 '18 at 1:20









rakaraka

556




556




closed as unclear what you're asking by Did, RRL, José Carlos Santos, Leucippus, Cesareo Dec 3 '18 at 10:02


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






closed as unclear what you're asking by Did, RRL, José Carlos Santos, Leucippus, Cesareo Dec 3 '18 at 10:02


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • $begingroup$
    WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
    $endgroup$
    – RRL
    Dec 3 '18 at 1:33










  • $begingroup$
    Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
    $endgroup$
    – raka
    Dec 3 '18 at 1:35






  • 3




    $begingroup$
    It is not bounded near $0$
    $endgroup$
    – RRL
    Dec 3 '18 at 1:37










  • $begingroup$
    I see it now. Thank you.
    $endgroup$
    – raka
    Dec 3 '18 at 1:38


















  • $begingroup$
    WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
    $endgroup$
    – RRL
    Dec 3 '18 at 1:33










  • $begingroup$
    Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
    $endgroup$
    – raka
    Dec 3 '18 at 1:35






  • 3




    $begingroup$
    It is not bounded near $0$
    $endgroup$
    – RRL
    Dec 3 '18 at 1:37










  • $begingroup$
    I see it now. Thank you.
    $endgroup$
    – raka
    Dec 3 '18 at 1:38
















$begingroup$
WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
$endgroup$
– RRL
Dec 3 '18 at 1:33




$begingroup$
WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
$endgroup$
– RRL
Dec 3 '18 at 1:33












$begingroup$
Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
$endgroup$
– raka
Dec 3 '18 at 1:35




$begingroup$
Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
$endgroup$
– raka
Dec 3 '18 at 1:35




3




3




$begingroup$
It is not bounded near $0$
$endgroup$
– RRL
Dec 3 '18 at 1:37




$begingroup$
It is not bounded near $0$
$endgroup$
– RRL
Dec 3 '18 at 1:37












$begingroup$
I see it now. Thank you.
$endgroup$
– raka
Dec 3 '18 at 1:38




$begingroup$
I see it now. Thank you.
$endgroup$
– raka
Dec 3 '18 at 1:38










1 Answer
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Take $x={1over{2pi n}}$, $f'(x)=-2pi n$






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Take $x={1over{2pi n}}$, $f'(x)=-2pi n$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Take $x={1over{2pi n}}$, $f'(x)=-2pi n$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Take $x={1over{2pi n}}$, $f'(x)=-2pi n$






        share|cite|improve this answer









        $endgroup$



        Take $x={1over{2pi n}}$, $f'(x)=-2pi n$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 1:27









        Tsemo AristideTsemo Aristide

        57k11444




        57k11444















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