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











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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




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












  • 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.
    – Xander Henderson
    Nov 28 at 0:45















up vote
2
down vote

favorite












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




















  • 1




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












  • 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.
    – Xander Henderson
    Nov 28 at 0:45













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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















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






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edited Nov 28 at 0:49









Servaes

22.2k33793




22.2k33793










asked Nov 28 at 0:41









Squaring the Circle is Easy

585




585








  • 1




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












  • 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.
    – Xander Henderson
    Nov 28 at 0:45














  • 1




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












  • 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.
    – Xander Henderson
    Nov 28 at 0:45








1




1




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






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














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.
– Xander Henderson
Nov 28 at 0:45




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.
– Xander Henderson
Nov 28 at 0:45










4 Answers
4






active

oldest

votes

















up vote
3
down vote



accepted










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



















  • 1




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










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


















up vote
1
down vote













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




























    up vote
    1
    down vote













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






    share|cite|improve this answer




























      up vote
      1
      down vote













      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





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        3
        down vote



        accepted










        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



















        • 1




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










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















        up vote
        3
        down vote



        accepted










        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



















        • 1




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










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













        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        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














        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 at 0:51

























        answered Nov 28 at 0:45









        Servaes

        22.2k33793




        22.2k33793








        • 1




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










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














        • 1




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










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








        1




        1




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




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












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




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










        up vote
        1
        down vote













        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

























          up vote
          1
          down vote













          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























            up vote
            1
            down vote










            up vote
            1
            down vote









            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












            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 at 0:45









            Siong Thye Goh

            97.7k1463116




            97.7k1463116






















                up vote
                1
                down vote













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






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













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






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









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






                    share|cite|improve this answer












                    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 at 0:45









                    Ethan Bolker

                    40.6k545107




                    40.6k545107






















                        up vote
                        1
                        down vote













                        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

























                          up vote
                          1
                          down vote













                          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























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            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












                            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 at 0:47









                            Y. Forman

                            11.4k523




                            11.4k523






























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