right derivative vs a right limit of the derivative times smallness factor












0












$begingroup$


In the question



Question on one-sided derivatives,



it is shown that if $f$ is differentiable on
$]x_0,x_0+delta[$ for some $delta>0$, such that
$;lim_{xrightarrow {x_{0}}^{+}}f^{prime}(x);$ exists, then
$$f^{prime}_{+}(x_{0})=lim_{epsilonrightarrow 0^{+}}
frac{f(x_0+epsilon)-f(x_0)}{epsilon}$$
exists, but the converse is not true.



Question:
Suppose $f$ is differentiable on
$]x_0,x_0+delta[$ for some $delta>0$.
Can $f^{prime}_{+}(x_{0})$ exist but the limit
$;lim_{xrightarrow {x_{0}}^{+}}(x-x_{0})f^{prime}(x);$ does not exist ?



In other words, does the existence of
$f^{prime}_{+}(x_{0})$ imply the existence of the limit $;lim_{xrightarrow {x_{0}}^{+}}(x-x_{0})f^{prime}(x);$ ?






real-analysis calculus analysis derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sorry. You are right. Let me correct the question.
    $endgroup$
    – Medo
    Jan 19 at 15:06










  • $begingroup$
    Yes. This is a counterexample. Thanks a lot.
    $endgroup$
    – Medo
    Jan 19 at 17:39
















0












$begingroup$


In the question



Question on one-sided derivatives,



it is shown that if $f$ is differentiable on
$]x_0,x_0+delta[$ for some $delta>0$, such that
$;lim_{xrightarrow {x_{0}}^{+}}f^{prime}(x);$ exists, then
$$f^{prime}_{+}(x_{0})=lim_{epsilonrightarrow 0^{+}}
frac{f(x_0+epsilon)-f(x_0)}{epsilon}$$
exists, but the converse is not true.



Question:
Suppose $f$ is differentiable on
$]x_0,x_0+delta[$ for some $delta>0$.
Can $f^{prime}_{+}(x_{0})$ exist but the limit
$;lim_{xrightarrow {x_{0}}^{+}}(x-x_{0})f^{prime}(x);$ does not exist ?



In other words, does the existence of
$f^{prime}_{+}(x_{0})$ imply the existence of the limit $;lim_{xrightarrow {x_{0}}^{+}}(x-x_{0})f^{prime}(x);$ ?






real-analysis calculus analysis derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sorry. You are right. Let me correct the question.
    $endgroup$
    – Medo
    Jan 19 at 15:06










  • $begingroup$
    Yes. This is a counterexample. Thanks a lot.
    $endgroup$
    – Medo
    Jan 19 at 17:39














0












0








0





$begingroup$


In the question



Question on one-sided derivatives,



it is shown that if $f$ is differentiable on
$]x_0,x_0+delta[$ for some $delta>0$, such that
$;lim_{xrightarrow {x_{0}}^{+}}f^{prime}(x);$ exists, then
$$f^{prime}_{+}(x_{0})=lim_{epsilonrightarrow 0^{+}}
frac{f(x_0+epsilon)-f(x_0)}{epsilon}$$
exists, but the converse is not true.



Question:
Suppose $f$ is differentiable on
$]x_0,x_0+delta[$ for some $delta>0$.
Can $f^{prime}_{+}(x_{0})$ exist but the limit
$;lim_{xrightarrow {x_{0}}^{+}}(x-x_{0})f^{prime}(x);$ does not exist ?



In other words, does the existence of
$f^{prime}_{+}(x_{0})$ imply the existence of the limit $;lim_{xrightarrow {x_{0}}^{+}}(x-x_{0})f^{prime}(x);$ ?






real-analysis calculus analysis derivatives







share|cite|improve this question











$endgroup$




In the question



Question on one-sided derivatives,



it is shown that if $f$ is differentiable on
$]x_0,x_0+delta[$ for some $delta>0$, such that
$;lim_{xrightarrow {x_{0}}^{+}}f^{prime}(x);$ exists, then
$$f^{prime}_{+}(x_{0})=lim_{epsilonrightarrow 0^{+}}
frac{f(x_0+epsilon)-f(x_0)}{epsilon}$$
exists, but the converse is not true.



Question:
Suppose $f$ is differentiable on
$]x_0,x_0+delta[$ for some $delta>0$.
Can $f^{prime}_{+}(x_{0})$ exist but the limit
$;lim_{xrightarrow {x_{0}}^{+}}(x-x_{0})f^{prime}(x);$ does not exist ?



In other words, does the existence of
$f^{prime}_{+}(x_{0})$ imply the existence of the limit $;lim_{xrightarrow {x_{0}}^{+}}(x-x_{0})f^{prime}(x);$ ?






real-analysis calculus analysis derivatives




real-analysis calculus analysis derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 15:22









Martin Sleziak

44.8k10118272




44.8k10118272










asked Jan 19 at 13:45









MedoMedo

632214




632214












  • $begingroup$
    Sorry. You are right. Let me correct the question.
    $endgroup$
    – Medo
    Jan 19 at 15:06










  • $begingroup$
    Yes. This is a counterexample. Thanks a lot.
    $endgroup$
    – Medo
    Jan 19 at 17:39


















  • $begingroup$
    Sorry. You are right. Let me correct the question.
    $endgroup$
    – Medo
    Jan 19 at 15:06










  • $begingroup$
    Yes. This is a counterexample. Thanks a lot.
    $endgroup$
    – Medo
    Jan 19 at 17:39
















$begingroup$
Sorry. You are right. Let me correct the question.
$endgroup$
– Medo
Jan 19 at 15:06




$begingroup$
Sorry. You are right. Let me correct the question.
$endgroup$
– Medo
Jan 19 at 15:06












$begingroup$
Yes. This is a counterexample. Thanks a lot.
$endgroup$
– Medo
Jan 19 at 17:39




$begingroup$
Yes. This is a counterexample. Thanks a lot.
$endgroup$
– Medo
Jan 19 at 17:39










1 Answer
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0












$begingroup$

Counterexample: $x_0=0$, $f(0)=0$, and
$$f(x)=x^2sin(x^{-3})$$
for $xne0$.






share|cite|improve this answer









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    0












    $begingroup$

    Counterexample: $x_0=0$, $f(0)=0$, and
    $$f(x)=x^2sin(x^{-3})$$
    for $xne0$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Counterexample: $x_0=0$, $f(0)=0$, and
      $$f(x)=x^2sin(x^{-3})$$
      for $xne0$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Counterexample: $x_0=0$, $f(0)=0$, and
        $$f(x)=x^2sin(x^{-3})$$
        for $xne0$.






        share|cite|improve this answer









        $endgroup$



        Counterexample: $x_0=0$, $f(0)=0$, and
        $$f(x)=x^2sin(x^{-3})$$
        for $xne0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 19 at 23:25









        bofbof

        51.8k558120




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