Is it true that $lim_{x to a}frac{f(x)-f(a)}{x-a}=pminfty implies lim_{x to a}f'(x) = pminfty$?












6












$begingroup$


Is it true that $lim_{x to a}frac{f(x)-f(a)}{x-a}=pminfty implies lim_{x to a}f'(x) = pminfty$?



Here, $f$ is a function defined on some open interval $I$, and $ain I$. Assume $f$ is continuous at $a$ and differentiable around $a$.



I can't for the life of me see how to provedisprove this implication, but my gut feeling is that it's false. Any guidance is greatly appreciated.






calculus







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  • $begingroup$
    isn't that definition of limit? the (ε, δ) should provide that the function is strictly increasing and must go unbounded to $ infty $
    $endgroup$
    – user29418
    Jan 7 at 1:28










  • $begingroup$
    Derivative of a differential function need not be continuous
    $endgroup$
    – Sorfosh
    Jan 7 at 6:00
















6












$begingroup$


Is it true that $lim_{x to a}frac{f(x)-f(a)}{x-a}=pminfty implies lim_{x to a}f'(x) = pminfty$?



Here, $f$ is a function defined on some open interval $I$, and $ain I$. Assume $f$ is continuous at $a$ and differentiable around $a$.



I can't for the life of me see how to provedisprove this implication, but my gut feeling is that it's false. Any guidance is greatly appreciated.






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    isn't that definition of limit? the (ε, δ) should provide that the function is strictly increasing and must go unbounded to $ infty $
    $endgroup$
    – user29418
    Jan 7 at 1:28










  • $begingroup$
    Derivative of a differential function need not be continuous
    $endgroup$
    – Sorfosh
    Jan 7 at 6:00














6












6








6


2



$begingroup$


Is it true that $lim_{x to a}frac{f(x)-f(a)}{x-a}=pminfty implies lim_{x to a}f'(x) = pminfty$?



Here, $f$ is a function defined on some open interval $I$, and $ain I$. Assume $f$ is continuous at $a$ and differentiable around $a$.



I can't for the life of me see how to provedisprove this implication, but my gut feeling is that it's false. Any guidance is greatly appreciated.






calculus







share|cite|improve this question











$endgroup$




Is it true that $lim_{x to a}frac{f(x)-f(a)}{x-a}=pminfty implies lim_{x to a}f'(x) = pminfty$?



Here, $f$ is a function defined on some open interval $I$, and $ain I$. Assume $f$ is continuous at $a$ and differentiable around $a$.



I can't for the life of me see how to provedisprove this implication, but my gut feeling is that it's false. Any guidance is greatly appreciated.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 1:14









amWhy

192k28225439




192k28225439










asked Jan 7 at 0:32









Euler's FriendEuler's Friend

415




415












  • $begingroup$
    isn't that definition of limit? the (ε, δ) should provide that the function is strictly increasing and must go unbounded to $ infty $
    $endgroup$
    – user29418
    Jan 7 at 1:28










  • $begingroup$
    Derivative of a differential function need not be continuous
    $endgroup$
    – Sorfosh
    Jan 7 at 6:00


















  • $begingroup$
    isn't that definition of limit? the (ε, δ) should provide that the function is strictly increasing and must go unbounded to $ infty $
    $endgroup$
    – user29418
    Jan 7 at 1:28










  • $begingroup$
    Derivative of a differential function need not be continuous
    $endgroup$
    – Sorfosh
    Jan 7 at 6:00
















$begingroup$
isn't that definition of limit? the (ε, δ) should provide that the function is strictly increasing and must go unbounded to $ infty $
$endgroup$
– user29418
Jan 7 at 1:28




$begingroup$
isn't that definition of limit? the (ε, δ) should provide that the function is strictly increasing and must go unbounded to $ infty $
$endgroup$
– user29418
Jan 7 at 1:28












$begingroup$
Derivative of a differential function need not be continuous
$endgroup$
– Sorfosh
Jan 7 at 6:00




$begingroup$
Derivative of a differential function need not be continuous
$endgroup$
– Sorfosh
Jan 7 at 6:00










1 Answer
1






active

oldest

votes


















8












$begingroup$

The claim is false. Consider
$$
g(x)=sqrt{x}sinfrac{1}{x}+x^{1/4},qquad x>0,
$$
and define $f$ on $mathbb R$ by
$$
f(x)=begin{cases}g(x) & text{for }x>0, \ 0 & text{for }x=0, \ -g(-x) & text{for }x<0.end{cases}
$$
Then $f$ is continuous everywhere, differentiable in $mathbb Rbackslash{0}$ and satisfies $f(-x)=-f(x)$.
Moreover, for $x>0$,
begin{align*}
frac{f(x)}{x}=frac{1}{sqrt{x}}sinfrac{1}{x}+frac{1}{x^{3/4}}geq-frac{1}{x^{1/2}}+frac{1}{x^{3/4}}longrightarrow+infty,quadtext{as }xto0+.
end{align*}
Due to the symmetry, the same is true for $x<0$ and $xto0-$. Thus, $f(x)/xto+infty$ as $xto0$.



Now, for $x>0$,
$$
f'(x)=frac{x^{3/4}+2 x sin left(frac{1}{x}right)-4 cos left(frac{1}{x}right)}{4 x^{3/2}}.
$$
However, the limit $lim_{xto0+}f'(x)$ does not even exist. For $x_n:=1/(npi)$ we have
begin{align*}
f'(x_n)=frac{pi ^{3/4}}{4 left(frac{1}{n}right)^{3/4}}-frac{pi ^{3/2} (-1)^n}{left(frac{1}{n}right)^{3/2}},
end{align*}
and so $f'(x_{2n})to-infty$, while $f'(x_{2n+1})to+infty$.






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






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    active

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    8












    $begingroup$

    The claim is false. Consider
    $$
    g(x)=sqrt{x}sinfrac{1}{x}+x^{1/4},qquad x>0,
    $$
    and define $f$ on $mathbb R$ by
    $$
    f(x)=begin{cases}g(x) & text{for }x>0, \ 0 & text{for }x=0, \ -g(-x) & text{for }x<0.end{cases}
    $$
    Then $f$ is continuous everywhere, differentiable in $mathbb Rbackslash{0}$ and satisfies $f(-x)=-f(x)$.
    Moreover, for $x>0$,
    begin{align*}
    frac{f(x)}{x}=frac{1}{sqrt{x}}sinfrac{1}{x}+frac{1}{x^{3/4}}geq-frac{1}{x^{1/2}}+frac{1}{x^{3/4}}longrightarrow+infty,quadtext{as }xto0+.
    end{align*}
    Due to the symmetry, the same is true for $x<0$ and $xto0-$. Thus, $f(x)/xto+infty$ as $xto0$.



    Now, for $x>0$,
    $$
    f'(x)=frac{x^{3/4}+2 x sin left(frac{1}{x}right)-4 cos left(frac{1}{x}right)}{4 x^{3/2}}.
    $$
    However, the limit $lim_{xto0+}f'(x)$ does not even exist. For $x_n:=1/(npi)$ we have
    begin{align*}
    f'(x_n)=frac{pi ^{3/4}}{4 left(frac{1}{n}right)^{3/4}}-frac{pi ^{3/2} (-1)^n}{left(frac{1}{n}right)^{3/2}},
    end{align*}
    and so $f'(x_{2n})to-infty$, while $f'(x_{2n+1})to+infty$.






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$

      The claim is false. Consider
      $$
      g(x)=sqrt{x}sinfrac{1}{x}+x^{1/4},qquad x>0,
      $$
      and define $f$ on $mathbb R$ by
      $$
      f(x)=begin{cases}g(x) & text{for }x>0, \ 0 & text{for }x=0, \ -g(-x) & text{for }x<0.end{cases}
      $$
      Then $f$ is continuous everywhere, differentiable in $mathbb Rbackslash{0}$ and satisfies $f(-x)=-f(x)$.
      Moreover, for $x>0$,
      begin{align*}
      frac{f(x)}{x}=frac{1}{sqrt{x}}sinfrac{1}{x}+frac{1}{x^{3/4}}geq-frac{1}{x^{1/2}}+frac{1}{x^{3/4}}longrightarrow+infty,quadtext{as }xto0+.
      end{align*}
      Due to the symmetry, the same is true for $x<0$ and $xto0-$. Thus, $f(x)/xto+infty$ as $xto0$.



      Now, for $x>0$,
      $$
      f'(x)=frac{x^{3/4}+2 x sin left(frac{1}{x}right)-4 cos left(frac{1}{x}right)}{4 x^{3/2}}.
      $$
      However, the limit $lim_{xto0+}f'(x)$ does not even exist. For $x_n:=1/(npi)$ we have
      begin{align*}
      f'(x_n)=frac{pi ^{3/4}}{4 left(frac{1}{n}right)^{3/4}}-frac{pi ^{3/2} (-1)^n}{left(frac{1}{n}right)^{3/2}},
      end{align*}
      and so $f'(x_{2n})to-infty$, while $f'(x_{2n+1})to+infty$.






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        The claim is false. Consider
        $$
        g(x)=sqrt{x}sinfrac{1}{x}+x^{1/4},qquad x>0,
        $$
        and define $f$ on $mathbb R$ by
        $$
        f(x)=begin{cases}g(x) & text{for }x>0, \ 0 & text{for }x=0, \ -g(-x) & text{for }x<0.end{cases}
        $$
        Then $f$ is continuous everywhere, differentiable in $mathbb Rbackslash{0}$ and satisfies $f(-x)=-f(x)$.
        Moreover, for $x>0$,
        begin{align*}
        frac{f(x)}{x}=frac{1}{sqrt{x}}sinfrac{1}{x}+frac{1}{x^{3/4}}geq-frac{1}{x^{1/2}}+frac{1}{x^{3/4}}longrightarrow+infty,quadtext{as }xto0+.
        end{align*}
        Due to the symmetry, the same is true for $x<0$ and $xto0-$. Thus, $f(x)/xto+infty$ as $xto0$.



        Now, for $x>0$,
        $$
        f'(x)=frac{x^{3/4}+2 x sin left(frac{1}{x}right)-4 cos left(frac{1}{x}right)}{4 x^{3/2}}.
        $$
        However, the limit $lim_{xto0+}f'(x)$ does not even exist. For $x_n:=1/(npi)$ we have
        begin{align*}
        f'(x_n)=frac{pi ^{3/4}}{4 left(frac{1}{n}right)^{3/4}}-frac{pi ^{3/2} (-1)^n}{left(frac{1}{n}right)^{3/2}},
        end{align*}
        and so $f'(x_{2n})to-infty$, while $f'(x_{2n+1})to+infty$.






        share|cite|improve this answer









        $endgroup$



        The claim is false. Consider
        $$
        g(x)=sqrt{x}sinfrac{1}{x}+x^{1/4},qquad x>0,
        $$
        and define $f$ on $mathbb R$ by
        $$
        f(x)=begin{cases}g(x) & text{for }x>0, \ 0 & text{for }x=0, \ -g(-x) & text{for }x<0.end{cases}
        $$
        Then $f$ is continuous everywhere, differentiable in $mathbb Rbackslash{0}$ and satisfies $f(-x)=-f(x)$.
        Moreover, for $x>0$,
        begin{align*}
        frac{f(x)}{x}=frac{1}{sqrt{x}}sinfrac{1}{x}+frac{1}{x^{3/4}}geq-frac{1}{x^{1/2}}+frac{1}{x^{3/4}}longrightarrow+infty,quadtext{as }xto0+.
        end{align*}
        Due to the symmetry, the same is true for $x<0$ and $xto0-$. Thus, $f(x)/xto+infty$ as $xto0$.



        Now, for $x>0$,
        $$
        f'(x)=frac{x^{3/4}+2 x sin left(frac{1}{x}right)-4 cos left(frac{1}{x}right)}{4 x^{3/2}}.
        $$
        However, the limit $lim_{xto0+}f'(x)$ does not even exist. For $x_n:=1/(npi)$ we have
        begin{align*}
        f'(x_n)=frac{pi ^{3/4}}{4 left(frac{1}{n}right)^{3/4}}-frac{pi ^{3/2} (-1)^n}{left(frac{1}{n}right)^{3/2}},
        end{align*}
        and so $f'(x_{2n})to-infty$, while $f'(x_{2n+1})to+infty$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 2:27









        sranthropsranthrop

        7,0241925




        7,0241925






























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