determine if the statement with limit is true












1












$begingroup$


I want to determine if the statement
$$lim_{xto a} f(x) = infty Rightarrow lim_{xto a}frac{1}{f(x)} = 0$$



is true or not (by proving it or proving a contradiction).



I know that I have a definition of limits that could be of help.
That is, if the function has a limit $A$ when $x rightarrow a$ then there is a number $ε > 0$ and $w$ such that



$$|x-a|< w Rightarrow |f(x)-A| < ε . $$



And I have all the other rules for limits. But I don't know how to go about this problem.






limits analysis proof-writing







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I want to determine if the statement
    $$lim_{xto a} f(x) = infty Rightarrow lim_{xto a}frac{1}{f(x)} = 0$$



    is true or not (by proving it or proving a contradiction).



    I know that I have a definition of limits that could be of help.
    That is, if the function has a limit $A$ when $x rightarrow a$ then there is a number $ε > 0$ and $w$ such that



    $$|x-a|< w Rightarrow |f(x)-A| < ε . $$



    And I have all the other rules for limits. But I don't know how to go about this problem.






    limits analysis proof-writing







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I want to determine if the statement
      $$lim_{xto a} f(x) = infty Rightarrow lim_{xto a}frac{1}{f(x)} = 0$$



      is true or not (by proving it or proving a contradiction).



      I know that I have a definition of limits that could be of help.
      That is, if the function has a limit $A$ when $x rightarrow a$ then there is a number $ε > 0$ and $w$ such that



      $$|x-a|< w Rightarrow |f(x)-A| < ε . $$



      And I have all the other rules for limits. But I don't know how to go about this problem.






      limits analysis proof-writing







      share|cite|improve this question











      $endgroup$




      I want to determine if the statement
      $$lim_{xto a} f(x) = infty Rightarrow lim_{xto a}frac{1}{f(x)} = 0$$



      is true or not (by proving it or proving a contradiction).



      I know that I have a definition of limits that could be of help.
      That is, if the function has a limit $A$ when $x rightarrow a$ then there is a number $ε > 0$ and $w$ such that



      $$|x-a|< w Rightarrow |f(x)-A| < ε . $$



      And I have all the other rules for limits. But I don't know how to go about this problem.






      limits analysis proof-writing




      limits analysis proof-writing






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 23 at 22:02









      J. W. Tanner

      3,2401320




      3,2401320










      asked Jan 23 at 19:02









      F WiF Wi

      384




      384






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          Alternatively we may use Heine's definition of the limit of a sequence.

          Let $x_n$ be a sequence so that $x_n rightarrow a => lim_{nto infty} f(x_n)=infty$.
          We have: $lim_{xto a} frac{1}{f(x)}=lim_{nto infty} frac{1}{f(x_n)}=frac{1}{lim_{ntoinfty} f(x_n)}=frac{1}{infty}=0$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            HINT



            We have to show that for each $epsilon > 0 exists delta > 0$ such that whenever $|x-a| < delta$ you have
            $$
            epsilon
            > left|frac{1}{f(x)} - 0 right|
            = left|frac1{f(x)}right|
            iff |f(x)| > 1/epsilon.
            $$



            Assume that as $x to a$, we have $f(x) to infty$, in other words, that $forall N > 0 exists delta>0$ such that $f(x) > N$ whenever $|x - a| < delta$.



            Can you see how to pick $epsilon = epsilon(N)$ to make what you want happen?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
              $endgroup$
              – F Wi
              Jan 23 at 22:01










            • $begingroup$
              @FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
              $endgroup$
              – gt6989b
              Jan 24 at 4:38



















            0












            $begingroup$

            We say that $lim_{xto a} f(x) = infty$, if $forall M>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $f(x) > M $.



            We say that $lim_{xto a}frac{1}{f(x)} = 0$, if $forall ε>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $left|frac{1}{f(x)} - 0 right| <epsilon$.



            It suffices to choose $epsilon = frac{1}{M}$, in which case $left|frac{1}{f(x)} - 0 right| <epsilon$ holds.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Alternatively we may use Heine's definition of the limit of a sequence.

              Let $x_n$ be a sequence so that $x_n rightarrow a => lim_{nto infty} f(x_n)=infty$.
              We have: $lim_{xto a} frac{1}{f(x)}=lim_{nto infty} frac{1}{f(x_n)}=frac{1}{lim_{ntoinfty} f(x_n)}=frac{1}{infty}=0$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Alternatively we may use Heine's definition of the limit of a sequence.

                Let $x_n$ be a sequence so that $x_n rightarrow a => lim_{nto infty} f(x_n)=infty$.
                We have: $lim_{xto a} frac{1}{f(x)}=lim_{nto infty} frac{1}{f(x_n)}=frac{1}{lim_{ntoinfty} f(x_n)}=frac{1}{infty}=0$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Alternatively we may use Heine's definition of the limit of a sequence.

                  Let $x_n$ be a sequence so that $x_n rightarrow a => lim_{nto infty} f(x_n)=infty$.
                  We have: $lim_{xto a} frac{1}{f(x)}=lim_{nto infty} frac{1}{f(x_n)}=frac{1}{lim_{ntoinfty} f(x_n)}=frac{1}{infty}=0$






                  share|cite|improve this answer









                  $endgroup$



                  Alternatively we may use Heine's definition of the limit of a sequence.

                  Let $x_n$ be a sequence so that $x_n rightarrow a => lim_{nto infty} f(x_n)=infty$.
                  We have: $lim_{xto a} frac{1}{f(x)}=lim_{nto infty} frac{1}{f(x_n)}=frac{1}{lim_{ntoinfty} f(x_n)}=frac{1}{infty}=0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 24 at 12:30









                  AlexdanutAlexdanut

                  1438




                  1438























                      0












                      $begingroup$

                      HINT



                      We have to show that for each $epsilon > 0 exists delta > 0$ such that whenever $|x-a| < delta$ you have
                      $$
                      epsilon
                      > left|frac{1}{f(x)} - 0 right|
                      = left|frac1{f(x)}right|
                      iff |f(x)| > 1/epsilon.
                      $$



                      Assume that as $x to a$, we have $f(x) to infty$, in other words, that $forall N > 0 exists delta>0$ such that $f(x) > N$ whenever $|x - a| < delta$.



                      Can you see how to pick $epsilon = epsilon(N)$ to make what you want happen?






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
                        $endgroup$
                        – F Wi
                        Jan 23 at 22:01










                      • $begingroup$
                        @FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
                        $endgroup$
                        – gt6989b
                        Jan 24 at 4:38
















                      0












                      $begingroup$

                      HINT



                      We have to show that for each $epsilon > 0 exists delta > 0$ such that whenever $|x-a| < delta$ you have
                      $$
                      epsilon
                      > left|frac{1}{f(x)} - 0 right|
                      = left|frac1{f(x)}right|
                      iff |f(x)| > 1/epsilon.
                      $$



                      Assume that as $x to a$, we have $f(x) to infty$, in other words, that $forall N > 0 exists delta>0$ such that $f(x) > N$ whenever $|x - a| < delta$.



                      Can you see how to pick $epsilon = epsilon(N)$ to make what you want happen?






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
                        $endgroup$
                        – F Wi
                        Jan 23 at 22:01










                      • $begingroup$
                        @FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
                        $endgroup$
                        – gt6989b
                        Jan 24 at 4:38














                      0












                      0








                      0





                      $begingroup$

                      HINT



                      We have to show that for each $epsilon > 0 exists delta > 0$ such that whenever $|x-a| < delta$ you have
                      $$
                      epsilon
                      > left|frac{1}{f(x)} - 0 right|
                      = left|frac1{f(x)}right|
                      iff |f(x)| > 1/epsilon.
                      $$



                      Assume that as $x to a$, we have $f(x) to infty$, in other words, that $forall N > 0 exists delta>0$ such that $f(x) > N$ whenever $|x - a| < delta$.



                      Can you see how to pick $epsilon = epsilon(N)$ to make what you want happen?






                      share|cite|improve this answer









                      $endgroup$



                      HINT



                      We have to show that for each $epsilon > 0 exists delta > 0$ such that whenever $|x-a| < delta$ you have
                      $$
                      epsilon
                      > left|frac{1}{f(x)} - 0 right|
                      = left|frac1{f(x)}right|
                      iff |f(x)| > 1/epsilon.
                      $$



                      Assume that as $x to a$, we have $f(x) to infty$, in other words, that $forall N > 0 exists delta>0$ such that $f(x) > N$ whenever $|x - a| < delta$.



                      Can you see how to pick $epsilon = epsilon(N)$ to make what you want happen?







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 23 at 19:10









                      gt6989bgt6989b

                      34.9k22557




                      34.9k22557












                      • $begingroup$
                        I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
                        $endgroup$
                        – F Wi
                        Jan 23 at 22:01










                      • $begingroup$
                        @FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
                        $endgroup$
                        – gt6989b
                        Jan 24 at 4:38


















                      • $begingroup$
                        I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
                        $endgroup$
                        – F Wi
                        Jan 23 at 22:01










                      • $begingroup$
                        @FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
                        $endgroup$
                        – gt6989b
                        Jan 24 at 4:38
















                      $begingroup$
                      I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
                      $endgroup$
                      – F Wi
                      Jan 23 at 22:01




                      $begingroup$
                      I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
                      $endgroup$
                      – F Wi
                      Jan 23 at 22:01












                      $begingroup$
                      @FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
                      $endgroup$
                      – gt6989b
                      Jan 24 at 4:38




                      $begingroup$
                      @FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
                      $endgroup$
                      – gt6989b
                      Jan 24 at 4:38











                      0












                      $begingroup$

                      We say that $lim_{xto a} f(x) = infty$, if $forall M>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $f(x) > M $.



                      We say that $lim_{xto a}frac{1}{f(x)} = 0$, if $forall ε>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $left|frac{1}{f(x)} - 0 right| <epsilon$.



                      It suffices to choose $epsilon = frac{1}{M}$, in which case $left|frac{1}{f(x)} - 0 right| <epsilon$ holds.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        We say that $lim_{xto a} f(x) = infty$, if $forall M>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $f(x) > M $.



                        We say that $lim_{xto a}frac{1}{f(x)} = 0$, if $forall ε>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $left|frac{1}{f(x)} - 0 right| <epsilon$.



                        It suffices to choose $epsilon = frac{1}{M}$, in which case $left|frac{1}{f(x)} - 0 right| <epsilon$ holds.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          We say that $lim_{xto a} f(x) = infty$, if $forall M>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $f(x) > M $.



                          We say that $lim_{xto a}frac{1}{f(x)} = 0$, if $forall ε>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $left|frac{1}{f(x)} - 0 right| <epsilon$.



                          It suffices to choose $epsilon = frac{1}{M}$, in which case $left|frac{1}{f(x)} - 0 right| <epsilon$ holds.






                          share|cite|improve this answer









                          $endgroup$



                          We say that $lim_{xto a} f(x) = infty$, if $forall M>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $f(x) > M $.



                          We say that $lim_{xto a}frac{1}{f(x)} = 0$, if $forall ε>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $left|frac{1}{f(x)} - 0 right| <epsilon$.



                          It suffices to choose $epsilon = frac{1}{M}$, in which case $left|frac{1}{f(x)} - 0 right| <epsilon$ holds.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 23 at 19:43









                          AlexandrosAlexandros

                          9401412




                          9401412






























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