$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$












1












$begingroup$


Suppose
$,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.



$$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$



My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.



If the statement is true could you either:




  • proof it

  • explain your intuition why this is true


     ... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?



If the statement is false could you either:




  • hint me a counter-example

  • disproof the statement


     ... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.


Cheers,

Pascal






limits







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Suppose
    $,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.



    $$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$



    My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.



    If the statement is true could you either:




    • proof it

    • explain your intuition why this is true


       ... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?



    If the statement is false could you either:




    • hint me a counter-example

    • disproof the statement


       ... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.


    Cheers,

    Pascal






    limits







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Suppose
      $,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.



      $$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$



      My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.



      If the statement is true could you either:




      • proof it

      • explain your intuition why this is true


         ... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?



      If the statement is false could you either:




      • hint me a counter-example

      • disproof the statement


         ... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.


      Cheers,

      Pascal






      limits







      share|cite|improve this question









      $endgroup$




      Suppose
      $,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.



      $$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$



      My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.



      If the statement is true could you either:




      • proof it

      • explain your intuition why this is true


         ... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?



      If the statement is false could you either:




      • hint me a counter-example

      • disproof the statement


         ... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.


      Cheers,

      Pascal






      limits




      limits






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 22:13









      plauerplauer

      355




      355






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          say $c$ is a finite real number then
          $limlimits_{xdownarrow0},f(x) = c $ means :



          $forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $



          meaning



          $forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $



          meaning



          $forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $



          meaning



          $limlimits_{xrightarrowinfty},f(frac1x) = c$



          if $c$ is $pm infty$ it's the same reasoning






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            It is true.



            First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.



            Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.



            Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?






            share|cite|improve this answer









            $endgroup$














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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

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              3












              $begingroup$

              say $c$ is a finite real number then
              $limlimits_{xdownarrow0},f(x) = c $ means :



              $forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $



              meaning



              $forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $



              meaning



              $forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $



              meaning



              $limlimits_{xrightarrowinfty},f(frac1x) = c$



              if $c$ is $pm infty$ it's the same reasoning






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                say $c$ is a finite real number then
                $limlimits_{xdownarrow0},f(x) = c $ means :



                $forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $



                meaning



                $forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $



                meaning



                $forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $



                meaning



                $limlimits_{xrightarrowinfty},f(frac1x) = c$



                if $c$ is $pm infty$ it's the same reasoning






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  say $c$ is a finite real number then
                  $limlimits_{xdownarrow0},f(x) = c $ means :



                  $forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $



                  meaning



                  $forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $



                  meaning



                  $forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $



                  meaning



                  $limlimits_{xrightarrowinfty},f(frac1x) = c$



                  if $c$ is $pm infty$ it's the same reasoning






                  share|cite|improve this answer









                  $endgroup$



                  say $c$ is a finite real number then
                  $limlimits_{xdownarrow0},f(x) = c $ means :



                  $forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $



                  meaning



                  $forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $



                  meaning



                  $forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $



                  meaning



                  $limlimits_{xrightarrowinfty},f(frac1x) = c$



                  if $c$ is $pm infty$ it's the same reasoning







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 18 '18 at 22:23









                  rapidracimrapidracim

                  1,7291419




                  1,7291419























                      0












                      $begingroup$

                      It is true.



                      First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.



                      Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.



                      Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        It is true.



                        First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.



                        Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.



                        Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          It is true.



                          First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.



                          Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.



                          Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?






                          share|cite|improve this answer









                          $endgroup$



                          It is true.



                          First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.



                          Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.



                          Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 18 '18 at 22:27









                          MikeMike

                          4,546512




                          4,546512






























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