existence of limit when both RHL and LHL approachees Positive infinity












1












$begingroup$


We know that if $$lim_{x to a^+}f(x)=lim_{x to a^-}f(x)=L$$ Then $$lim_{x to a}f(x) =L$$ if $L$ is finite



But if $$lim_{x to a^+}f(x) to +infty$$ and
$$lim_{x to a^-}f(x) to +infty$$



Can we say $$lim_{x to a} f(x)$$ Does not exists since we cannot compare two infinities.






calculus limits







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$endgroup$












  • $begingroup$
    What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
    $endgroup$
    – T. Bongers
    Jan 5 '16 at 4:17










  • $begingroup$
    what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
    $endgroup$
    – Umesh shankar
    Jan 5 '16 at 4:19






  • 1




    $begingroup$
    No, the limit is equal to infinity. How would subtraction be involved?
    $endgroup$
    – T. Bongers
    Jan 5 '16 at 4:20


















1












$begingroup$


We know that if $$lim_{x to a^+}f(x)=lim_{x to a^-}f(x)=L$$ Then $$lim_{x to a}f(x) =L$$ if $L$ is finite



But if $$lim_{x to a^+}f(x) to +infty$$ and
$$lim_{x to a^-}f(x) to +infty$$



Can we say $$lim_{x to a} f(x)$$ Does not exists since we cannot compare two infinities.






calculus limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
    $endgroup$
    – T. Bongers
    Jan 5 '16 at 4:17










  • $begingroup$
    what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
    $endgroup$
    – Umesh shankar
    Jan 5 '16 at 4:19






  • 1




    $begingroup$
    No, the limit is equal to infinity. How would subtraction be involved?
    $endgroup$
    – T. Bongers
    Jan 5 '16 at 4:20
















1












1








1





$begingroup$


We know that if $$lim_{x to a^+}f(x)=lim_{x to a^-}f(x)=L$$ Then $$lim_{x to a}f(x) =L$$ if $L$ is finite



But if $$lim_{x to a^+}f(x) to +infty$$ and
$$lim_{x to a^-}f(x) to +infty$$



Can we say $$lim_{x to a} f(x)$$ Does not exists since we cannot compare two infinities.






calculus limits







share|cite|improve this question









$endgroup$




We know that if $$lim_{x to a^+}f(x)=lim_{x to a^-}f(x)=L$$ Then $$lim_{x to a}f(x) =L$$ if $L$ is finite



But if $$lim_{x to a^+}f(x) to +infty$$ and
$$lim_{x to a^-}f(x) to +infty$$



Can we say $$lim_{x to a} f(x)$$ Does not exists since we cannot compare two infinities.






calculus limits




calculus limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 '16 at 4:15









Umesh shankarUmesh shankar

3,09231220




3,09231220












  • $begingroup$
    What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
    $endgroup$
    – T. Bongers
    Jan 5 '16 at 4:17










  • $begingroup$
    what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
    $endgroup$
    – Umesh shankar
    Jan 5 '16 at 4:19






  • 1




    $begingroup$
    No, the limit is equal to infinity. How would subtraction be involved?
    $endgroup$
    – T. Bongers
    Jan 5 '16 at 4:20




















  • $begingroup$
    What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
    $endgroup$
    – T. Bongers
    Jan 5 '16 at 4:17










  • $begingroup$
    what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
    $endgroup$
    – Umesh shankar
    Jan 5 '16 at 4:19






  • 1




    $begingroup$
    No, the limit is equal to infinity. How would subtraction be involved?
    $endgroup$
    – T. Bongers
    Jan 5 '16 at 4:20


















$begingroup$
What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
$endgroup$
– T. Bongers
Jan 5 '16 at 4:17




$begingroup$
What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
$endgroup$
– T. Bongers
Jan 5 '16 at 4:17












$begingroup$
what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
$endgroup$
– Umesh shankar
Jan 5 '16 at 4:19




$begingroup$
what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
$endgroup$
– Umesh shankar
Jan 5 '16 at 4:19




1




1




$begingroup$
No, the limit is equal to infinity. How would subtraction be involved?
$endgroup$
– T. Bongers
Jan 5 '16 at 4:20






$begingroup$
No, the limit is equal to infinity. How would subtraction be involved?
$endgroup$
– T. Bongers
Jan 5 '16 at 4:20












3 Answers
3






active

oldest

votes


















0












$begingroup$

Having $$lim_{x to a^+} f(x)=infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$lim_{x to a} f(x)=L,$$ which implies that $$lim_{x to a^+} f(x)=L.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Limits of $pm infty$ can exist.
    $endgroup$
    – zhw.
    May 14 '17 at 21:39



















0












$begingroup$

In general, we only say a limit 'exists' when it is finite. When we say $lim_{xto a}f(x)=+infty$ we mean that



$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta),, f(x)>M tag{1}$$



In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.




When we say $lim_{xto a^+}f(x)=+infty$ we mean that



$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x>a),, f(x)>Mtag{2}$$



In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.




When we say $lim_{xto a^-}f(x)=+infty$ we mean that



$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x<a),, f(x)>Mtag{3}$$



In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.




Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$


    Can we say $lim limits_{xrightarrow a}f(x)$ does not exists since we cannot compare two infinities?




    Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $pminfty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Having $$lim_{x to a^+} f(x)=infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$lim_{x to a} f(x)=L,$$ which implies that $$lim_{x to a^+} f(x)=L.$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Limits of $pm infty$ can exist.
        $endgroup$
        – zhw.
        May 14 '17 at 21:39
















      0












      $begingroup$

      Having $$lim_{x to a^+} f(x)=infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$lim_{x to a} f(x)=L,$$ which implies that $$lim_{x to a^+} f(x)=L.$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Limits of $pm infty$ can exist.
        $endgroup$
        – zhw.
        May 14 '17 at 21:39














      0












      0








      0





      $begingroup$

      Having $$lim_{x to a^+} f(x)=infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$lim_{x to a} f(x)=L,$$ which implies that $$lim_{x to a^+} f(x)=L.$$






      share|cite|improve this answer









      $endgroup$



      Having $$lim_{x to a^+} f(x)=infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$lim_{x to a} f(x)=L,$$ which implies that $$lim_{x to a^+} f(x)=L.$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 5 '16 at 4:36









      zwischzwisch

      12




      12












      • $begingroup$
        Limits of $pm infty$ can exist.
        $endgroup$
        – zhw.
        May 14 '17 at 21:39


















      • $begingroup$
        Limits of $pm infty$ can exist.
        $endgroup$
        – zhw.
        May 14 '17 at 21:39
















      $begingroup$
      Limits of $pm infty$ can exist.
      $endgroup$
      – zhw.
      May 14 '17 at 21:39




      $begingroup$
      Limits of $pm infty$ can exist.
      $endgroup$
      – zhw.
      May 14 '17 at 21:39











      0












      $begingroup$

      In general, we only say a limit 'exists' when it is finite. When we say $lim_{xto a}f(x)=+infty$ we mean that



      $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta),, f(x)>M tag{1}$$



      In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.




      When we say $lim_{xto a^+}f(x)=+infty$ we mean that



      $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x>a),, f(x)>Mtag{2}$$



      In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.




      When we say $lim_{xto a^-}f(x)=+infty$ we mean that



      $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x<a),, f(x)>Mtag{3}$$



      In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.




      Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        In general, we only say a limit 'exists' when it is finite. When we say $lim_{xto a}f(x)=+infty$ we mean that



        $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta),, f(x)>M tag{1}$$



        In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.




        When we say $lim_{xto a^+}f(x)=+infty$ we mean that



        $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x>a),, f(x)>Mtag{2}$$



        In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.




        When we say $lim_{xto a^-}f(x)=+infty$ we mean that



        $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x<a),, f(x)>Mtag{3}$$



        In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.




        Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          In general, we only say a limit 'exists' when it is finite. When we say $lim_{xto a}f(x)=+infty$ we mean that



          $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta),, f(x)>M tag{1}$$



          In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.




          When we say $lim_{xto a^+}f(x)=+infty$ we mean that



          $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x>a),, f(x)>Mtag{2}$$



          In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.




          When we say $lim_{xto a^-}f(x)=+infty$ we mean that



          $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x<a),, f(x)>Mtag{3}$$



          In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.




          Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?






          share|cite|improve this answer









          $endgroup$



          In general, we only say a limit 'exists' when it is finite. When we say $lim_{xto a}f(x)=+infty$ we mean that



          $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta),, f(x)>M tag{1}$$



          In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.




          When we say $lim_{xto a^+}f(x)=+infty$ we mean that



          $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x>a),, f(x)>Mtag{2}$$



          In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.




          When we say $lim_{xto a^-}f(x)=+infty$ we mean that



          $$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x<a),, f(x)>Mtag{3}$$



          In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.




          Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 6 '17 at 14:18









          FimpellizieriFimpellizieri

          17.2k11836




          17.2k11836























              0












              $begingroup$


              Can we say $lim limits_{xrightarrow a}f(x)$ does not exists since we cannot compare two infinities?




              Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $pminfty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$


                Can we say $lim limits_{xrightarrow a}f(x)$ does not exists since we cannot compare two infinities?




                Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $pminfty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$


                  Can we say $lim limits_{xrightarrow a}f(x)$ does not exists since we cannot compare two infinities?




                  Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $pminfty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.






                  share|cite|improve this answer











                  $endgroup$




                  Can we say $lim limits_{xrightarrow a}f(x)$ does not exists since we cannot compare two infinities?




                  Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $pminfty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 6 '17 at 14:40

























                  answered Nov 6 '17 at 14:34









                  Michael RybkinMichael Rybkin

                  4,244422




                  4,244422






























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