Example showing $limlimits_{x to x_0} xf(x) neq x_0limlimits_{x to x_0} f(x)$












2












$begingroup$


I can looking for a simple example to illustrate $limlimits_{x to x_0} xf(x) neq x_0 limlimits_{x to x_0} f(x)$



For example I have tried $f(x) = x-1, x_0 = 1$ hoping that I would get a zero on one side and a non-zero on the other, but so far without success.



Can someone provide an example to this statement?






calculus algebra-precalculus limits examples-counterexamples discontinuous-functions







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  • 1




    $begingroup$
    $lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
    $endgroup$
    – AlkaKadri
    Nov 28 '18 at 0:45












  • $begingroup$
    Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
    $endgroup$
    – Xander Henderson
    Nov 28 '18 at 0:45
















2












$begingroup$


I can looking for a simple example to illustrate $limlimits_{x to x_0} xf(x) neq x_0 limlimits_{x to x_0} f(x)$



For example I have tried $f(x) = x-1, x_0 = 1$ hoping that I would get a zero on one side and a non-zero on the other, but so far without success.



Can someone provide an example to this statement?






calculus algebra-precalculus limits examples-counterexamples discontinuous-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
    $endgroup$
    – AlkaKadri
    Nov 28 '18 at 0:45












  • $begingroup$
    Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
    $endgroup$
    – Xander Henderson
    Nov 28 '18 at 0:45














2












2








2





$begingroup$


I can looking for a simple example to illustrate $limlimits_{x to x_0} xf(x) neq x_0 limlimits_{x to x_0} f(x)$



For example I have tried $f(x) = x-1, x_0 = 1$ hoping that I would get a zero on one side and a non-zero on the other, but so far without success.



Can someone provide an example to this statement?






calculus algebra-precalculus limits examples-counterexamples discontinuous-functions







share|cite|improve this question











$endgroup$




I can looking for a simple example to illustrate $limlimits_{x to x_0} xf(x) neq x_0 limlimits_{x to x_0} f(x)$



For example I have tried $f(x) = x-1, x_0 = 1$ hoping that I would get a zero on one side and a non-zero on the other, but so far without success.



Can someone provide an example to this statement?






calculus algebra-precalculus limits examples-counterexamples discontinuous-functions




calculus algebra-precalculus limits examples-counterexamples discontinuous-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 0:49









Servaes

23.4k33893




23.4k33893










asked Nov 28 '18 at 0:41









Squaring the Circle is EasySquaring the Circle is Easy

585




585








  • 1




    $begingroup$
    $lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
    $endgroup$
    – AlkaKadri
    Nov 28 '18 at 0:45












  • $begingroup$
    Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
    $endgroup$
    – Xander Henderson
    Nov 28 '18 at 0:45














  • 1




    $begingroup$
    $lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
    $endgroup$
    – AlkaKadri
    Nov 28 '18 at 0:45












  • $begingroup$
    Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
    $endgroup$
    – Xander Henderson
    Nov 28 '18 at 0:45








1




1




$begingroup$
$lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:45






$begingroup$
$lim_{x to 0} xfrac{1}{x}$, but if you insist that $lim_{x to x_o} f(x)$ needs to always be defined, then such an example doesn't exist
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:45














$begingroup$
Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
$endgroup$
– Xander Henderson
Nov 28 '18 at 0:45




$begingroup$
Take $f(x) = frac{1}{x}$ and $x_0 = 0$. Then $$lim_{xto x_0} x f(x) = lim_{xto 0} 1 = 1, $$ but $x_0 lim_{xto x_0} f(x)$ is not defined.
$endgroup$
– Xander Henderson
Nov 28 '18 at 0:45










4 Answers
4






active

oldest

votes


















3












$begingroup$

HINT: If $lim_{xto x_0}f(x)$ exists then by the product rule for limits
$$lim_{xto x_0}xf(x)
=left(lim_{xto x_0}xright)left(lim_{xto x_0}f(x)right)
=x_0lim_{xto x_0}f(x),$$
so you want to find some function $f$ and some point $x_0$ such that $lim_{xto x_0}f(x)$ does not exist.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Existence of the limit is enough for the product rule; continuity is not required.
    $endgroup$
    – Y. Forman
    Nov 28 '18 at 0:49










  • $begingroup$
    @Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
    $endgroup$
    – Servaes
    Nov 28 '18 at 0:54



















1












$begingroup$

Let $f(x)=frac1x$, $x_0=0$, then on the LHS we have $1$.



On the right hand side $lim_{x to x_0} f(x)$ is not defined.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    How about $x_0 = 0$, $f(x) = 1/x^2$?






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      If both limits exist, the equality is true by the product rule of limits: $lim_{xto x_0} xf(x) = lim_{xto x_0} x lim _{xto x_0}f(x) = x_0 lim _{xto x_0}f(x)$



      So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=frac1x$ with $x_0=0$






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        HINT: If $lim_{xto x_0}f(x)$ exists then by the product rule for limits
        $$lim_{xto x_0}xf(x)
        =left(lim_{xto x_0}xright)left(lim_{xto x_0}f(x)right)
        =x_0lim_{xto x_0}f(x),$$
        so you want to find some function $f$ and some point $x_0$ such that $lim_{xto x_0}f(x)$ does not exist.






        share|cite|improve this answer











        $endgroup$









        • 1




          $begingroup$
          Existence of the limit is enough for the product rule; continuity is not required.
          $endgroup$
          – Y. Forman
          Nov 28 '18 at 0:49










        • $begingroup$
          @Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
          $endgroup$
          – Servaes
          Nov 28 '18 at 0:54
















        3












        $begingroup$

        HINT: If $lim_{xto x_0}f(x)$ exists then by the product rule for limits
        $$lim_{xto x_0}xf(x)
        =left(lim_{xto x_0}xright)left(lim_{xto x_0}f(x)right)
        =x_0lim_{xto x_0}f(x),$$
        so you want to find some function $f$ and some point $x_0$ such that $lim_{xto x_0}f(x)$ does not exist.






        share|cite|improve this answer











        $endgroup$









        • 1




          $begingroup$
          Existence of the limit is enough for the product rule; continuity is not required.
          $endgroup$
          – Y. Forman
          Nov 28 '18 at 0:49










        • $begingroup$
          @Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
          $endgroup$
          – Servaes
          Nov 28 '18 at 0:54














        3












        3








        3





        $begingroup$

        HINT: If $lim_{xto x_0}f(x)$ exists then by the product rule for limits
        $$lim_{xto x_0}xf(x)
        =left(lim_{xto x_0}xright)left(lim_{xto x_0}f(x)right)
        =x_0lim_{xto x_0}f(x),$$
        so you want to find some function $f$ and some point $x_0$ such that $lim_{xto x_0}f(x)$ does not exist.






        share|cite|improve this answer











        $endgroup$



        HINT: If $lim_{xto x_0}f(x)$ exists then by the product rule for limits
        $$lim_{xto x_0}xf(x)
        =left(lim_{xto x_0}xright)left(lim_{xto x_0}f(x)right)
        =x_0lim_{xto x_0}f(x),$$
        so you want to find some function $f$ and some point $x_0$ such that $lim_{xto x_0}f(x)$ does not exist.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 28 '18 at 0:51

























        answered Nov 28 '18 at 0:45









        ServaesServaes

        23.4k33893




        23.4k33893








        • 1




          $begingroup$
          Existence of the limit is enough for the product rule; continuity is not required.
          $endgroup$
          – Y. Forman
          Nov 28 '18 at 0:49










        • $begingroup$
          @Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
          $endgroup$
          – Servaes
          Nov 28 '18 at 0:54














        • 1




          $begingroup$
          Existence of the limit is enough for the product rule; continuity is not required.
          $endgroup$
          – Y. Forman
          Nov 28 '18 at 0:49










        • $begingroup$
          @Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
          $endgroup$
          – Servaes
          Nov 28 '18 at 0:54








        1




        1




        $begingroup$
        Existence of the limit is enough for the product rule; continuity is not required.
        $endgroup$
        – Y. Forman
        Nov 28 '18 at 0:49




        $begingroup$
        Existence of the limit is enough for the product rule; continuity is not required.
        $endgroup$
        – Y. Forman
        Nov 28 '18 at 0:49












        $begingroup$
        @Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
        $endgroup$
        – Servaes
        Nov 28 '18 at 0:54




        $begingroup$
        @Y.Forman I've adjusted my answer accordingly. It seems our answers have converged.
        $endgroup$
        – Servaes
        Nov 28 '18 at 0:54











        1












        $begingroup$

        Let $f(x)=frac1x$, $x_0=0$, then on the LHS we have $1$.



        On the right hand side $lim_{x to x_0} f(x)$ is not defined.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Let $f(x)=frac1x$, $x_0=0$, then on the LHS we have $1$.



          On the right hand side $lim_{x to x_0} f(x)$ is not defined.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Let $f(x)=frac1x$, $x_0=0$, then on the LHS we have $1$.



            On the right hand side $lim_{x to x_0} f(x)$ is not defined.






            share|cite|improve this answer









            $endgroup$



            Let $f(x)=frac1x$, $x_0=0$, then on the LHS we have $1$.



            On the right hand side $lim_{x to x_0} f(x)$ is not defined.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 28 '18 at 0:45









            Siong Thye GohSiong Thye Goh

            101k1466117




            101k1466117























                1












                $begingroup$

                How about $x_0 = 0$, $f(x) = 1/x^2$?






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  How about $x_0 = 0$, $f(x) = 1/x^2$?






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    How about $x_0 = 0$, $f(x) = 1/x^2$?






                    share|cite|improve this answer









                    $endgroup$



                    How about $x_0 = 0$, $f(x) = 1/x^2$?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 28 '18 at 0:45









                    Ethan BolkerEthan Bolker

                    42.4k549112




                    42.4k549112























                        1












                        $begingroup$

                        If both limits exist, the equality is true by the product rule of limits: $lim_{xto x_0} xf(x) = lim_{xto x_0} x lim _{xto x_0}f(x) = x_0 lim _{xto x_0}f(x)$



                        So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=frac1x$ with $x_0=0$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          If both limits exist, the equality is true by the product rule of limits: $lim_{xto x_0} xf(x) = lim_{xto x_0} x lim _{xto x_0}f(x) = x_0 lim _{xto x_0}f(x)$



                          So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=frac1x$ with $x_0=0$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            If both limits exist, the equality is true by the product rule of limits: $lim_{xto x_0} xf(x) = lim_{xto x_0} x lim _{xto x_0}f(x) = x_0 lim _{xto x_0}f(x)$



                            So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=frac1x$ with $x_0=0$






                            share|cite|improve this answer









                            $endgroup$



                            If both limits exist, the equality is true by the product rule of limits: $lim_{xto x_0} xf(x) = lim_{xto x_0} x lim _{xto x_0}f(x) = x_0 lim _{xto x_0}f(x)$



                            So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=frac1x$ with $x_0=0$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 28 '18 at 0:47









                            Y. FormanY. Forman

                            11.5k523




                            11.5k523






























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