How to apply algebra of limits?











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I am a bit confused about application of algebra of limits in solving problems.
I know that $lim f(x)g(x) = lim f(x) cdot lim g(x)$.



While solving problems on limits I partitioned the function in two such parts so that their individual limits can easily be found. But I didn't get right answer. (For example, I get a finite limit of first function, and limit of second part of function doesn't exist. But the limit of whole function exists finitely.)



So how to know if this algebra can be used to find a particular limit or not?






calculus







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  • 2




    You may use that $lim_{ntoinfty} a_ncdot b_n = lim_{ntoinfty} a_n cdot lim_{ntoinfty} b_n$ whenever $lim_{ntoinfty} a_n$ and $lim_{ntoinfty} b_n$ exist.
    – bubububub
    Nov 26 at 6:29

















up vote
0
down vote

favorite












I am a bit confused about application of algebra of limits in solving problems.
I know that $lim f(x)g(x) = lim f(x) cdot lim g(x)$.



While solving problems on limits I partitioned the function in two such parts so that their individual limits can easily be found. But I didn't get right answer. (For example, I get a finite limit of first function, and limit of second part of function doesn't exist. But the limit of whole function exists finitely.)



So how to know if this algebra can be used to find a particular limit or not?






calculus







share|cite|improve this question




















  • 2




    You may use that $lim_{ntoinfty} a_ncdot b_n = lim_{ntoinfty} a_n cdot lim_{ntoinfty} b_n$ whenever $lim_{ntoinfty} a_n$ and $lim_{ntoinfty} b_n$ exist.
    – bubububub
    Nov 26 at 6:29















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am a bit confused about application of algebra of limits in solving problems.
I know that $lim f(x)g(x) = lim f(x) cdot lim g(x)$.



While solving problems on limits I partitioned the function in two such parts so that their individual limits can easily be found. But I didn't get right answer. (For example, I get a finite limit of first function, and limit of second part of function doesn't exist. But the limit of whole function exists finitely.)



So how to know if this algebra can be used to find a particular limit or not?






calculus







share|cite|improve this question















I am a bit confused about application of algebra of limits in solving problems.
I know that $lim f(x)g(x) = lim f(x) cdot lim g(x)$.



While solving problems on limits I partitioned the function in two such parts so that their individual limits can easily be found. But I didn't get right answer. (For example, I get a finite limit of first function, and limit of second part of function doesn't exist. But the limit of whole function exists finitely.)



So how to know if this algebra can be used to find a particular limit or not?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 6:34









Eevee Trainer

2,629221




2,629221










asked Nov 26 at 6:27









Mathsaddict

858




858








  • 2




    You may use that $lim_{ntoinfty} a_ncdot b_n = lim_{ntoinfty} a_n cdot lim_{ntoinfty} b_n$ whenever $lim_{ntoinfty} a_n$ and $lim_{ntoinfty} b_n$ exist.
    – bubububub
    Nov 26 at 6:29
















  • 2




    You may use that $lim_{ntoinfty} a_ncdot b_n = lim_{ntoinfty} a_n cdot lim_{ntoinfty} b_n$ whenever $lim_{ntoinfty} a_n$ and $lim_{ntoinfty} b_n$ exist.
    – bubububub
    Nov 26 at 6:29










2




2




You may use that $lim_{ntoinfty} a_ncdot b_n = lim_{ntoinfty} a_n cdot lim_{ntoinfty} b_n$ whenever $lim_{ntoinfty} a_n$ and $lim_{ntoinfty} b_n$ exist.
– bubububub
Nov 26 at 6:29






You may use that $lim_{ntoinfty} a_ncdot b_n = lim_{ntoinfty} a_n cdot lim_{ntoinfty} b_n$ whenever $lim_{ntoinfty} a_n$ and $lim_{ntoinfty} b_n$ exist.
– bubububub
Nov 26 at 6:29












3 Answers
3






active

oldest

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up vote
1
down vote













We can use



$$lim_{xto x_0} f(x)cdot g(x)=lim_{xto x_0} f(x)cdot lim_{xto x_0} g(x)$$



when both limits exist finite otherwise we can also conclude directly when one is finite $Lneq0$ and the other infinite or when both are infinite using the rules





  • $L cdot infty= infty$

  • $infty cdot infty= infty$


selecting the sign accordingly.






share|cite|improve this answer




























    up vote
    1
    down vote













    The limit of the product of functions equals the product of the limits of the functions only if both of those latter limits converge to some finite number. That is to say,



    $$lim_{x to c} f(x)g(x) = lim_{x to c} f(x) cdot lim_{x to c} g(x)$$



    only if $lim_{x to c} f(x)$ and $lim_{x to c} g(x)$ are some finite number. (It is this existence of the limits part that is essential. There are ways to handle cases where $f,g$ approach infinity as well, but it's not through the above equality.)



    (A minor note: a lot of similar laws, invoking for example the sum and difference of functions, also rely on this premise.)






    share|cite|improve this answer























    • Fair point, I think I more or less corrected that nuance.
      – Eevee Trainer
      Nov 26 at 6:44










    • limit [(1 - sinx)/(pi - 2x)^4)(cosx)(8x^3 - pi^3)] as x tends to pi/2. Here the limit of (1-sinx)/(pi-2x)^4 is some non zero finite number. And the limit of rest part of function is 0. Then why the limit of the whole function is not zero?
      – Mathsaddict
      Nov 26 at 7:05


















    up vote
    0
    down vote













    The fact that you decomposed your function into a product with one undetermined factor does not question the truth of the statement all the other answers and comment point out. Just an unfortunate choice of writing your function. Let for instance the inverse function
    $$frac 1 x$$



    It converges, but if you write it as
    $$frac{e^{ix}}{x} e^{-ix}$$



    then the first factor converges to zero and the second one does not converge. The problem is not the algebras on limits (it applies whenever both limits do exist).






    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








      up vote
      1
      down vote













      We can use



      $$lim_{xto x_0} f(x)cdot g(x)=lim_{xto x_0} f(x)cdot lim_{xto x_0} g(x)$$



      when both limits exist finite otherwise we can also conclude directly when one is finite $Lneq0$ and the other infinite or when both are infinite using the rules





      • $L cdot infty= infty$

      • $infty cdot infty= infty$


      selecting the sign accordingly.






      share|cite|improve this answer

























        up vote
        1
        down vote













        We can use



        $$lim_{xto x_0} f(x)cdot g(x)=lim_{xto x_0} f(x)cdot lim_{xto x_0} g(x)$$



        when both limits exist finite otherwise we can also conclude directly when one is finite $Lneq0$ and the other infinite or when both are infinite using the rules





        • $L cdot infty= infty$

        • $infty cdot infty= infty$


        selecting the sign accordingly.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          We can use



          $$lim_{xto x_0} f(x)cdot g(x)=lim_{xto x_0} f(x)cdot lim_{xto x_0} g(x)$$



          when both limits exist finite otherwise we can also conclude directly when one is finite $Lneq0$ and the other infinite or when both are infinite using the rules





          • $L cdot infty= infty$

          • $infty cdot infty= infty$


          selecting the sign accordingly.






          share|cite|improve this answer












          We can use



          $$lim_{xto x_0} f(x)cdot g(x)=lim_{xto x_0} f(x)cdot lim_{xto x_0} g(x)$$



          when both limits exist finite otherwise we can also conclude directly when one is finite $Lneq0$ and the other infinite or when both are infinite using the rules





          • $L cdot infty= infty$

          • $infty cdot infty= infty$


          selecting the sign accordingly.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 6:35









          gimusi

          91.9k84495




          91.9k84495






















              up vote
              1
              down vote













              The limit of the product of functions equals the product of the limits of the functions only if both of those latter limits converge to some finite number. That is to say,



              $$lim_{x to c} f(x)g(x) = lim_{x to c} f(x) cdot lim_{x to c} g(x)$$



              only if $lim_{x to c} f(x)$ and $lim_{x to c} g(x)$ are some finite number. (It is this existence of the limits part that is essential. There are ways to handle cases where $f,g$ approach infinity as well, but it's not through the above equality.)



              (A minor note: a lot of similar laws, invoking for example the sum and difference of functions, also rely on this premise.)






              share|cite|improve this answer























              • Fair point, I think I more or less corrected that nuance.
                – Eevee Trainer
                Nov 26 at 6:44










              • limit [(1 - sinx)/(pi - 2x)^4)(cosx)(8x^3 - pi^3)] as x tends to pi/2. Here the limit of (1-sinx)/(pi-2x)^4 is some non zero finite number. And the limit of rest part of function is 0. Then why the limit of the whole function is not zero?
                – Mathsaddict
                Nov 26 at 7:05















              up vote
              1
              down vote













              The limit of the product of functions equals the product of the limits of the functions only if both of those latter limits converge to some finite number. That is to say,



              $$lim_{x to c} f(x)g(x) = lim_{x to c} f(x) cdot lim_{x to c} g(x)$$



              only if $lim_{x to c} f(x)$ and $lim_{x to c} g(x)$ are some finite number. (It is this existence of the limits part that is essential. There are ways to handle cases where $f,g$ approach infinity as well, but it's not through the above equality.)



              (A minor note: a lot of similar laws, invoking for example the sum and difference of functions, also rely on this premise.)






              share|cite|improve this answer























              • Fair point, I think I more or less corrected that nuance.
                – Eevee Trainer
                Nov 26 at 6:44










              • limit [(1 - sinx)/(pi - 2x)^4)(cosx)(8x^3 - pi^3)] as x tends to pi/2. Here the limit of (1-sinx)/(pi-2x)^4 is some non zero finite number. And the limit of rest part of function is 0. Then why the limit of the whole function is not zero?
                – Mathsaddict
                Nov 26 at 7:05













              up vote
              1
              down vote










              up vote
              1
              down vote









              The limit of the product of functions equals the product of the limits of the functions only if both of those latter limits converge to some finite number. That is to say,



              $$lim_{x to c} f(x)g(x) = lim_{x to c} f(x) cdot lim_{x to c} g(x)$$



              only if $lim_{x to c} f(x)$ and $lim_{x to c} g(x)$ are some finite number. (It is this existence of the limits part that is essential. There are ways to handle cases where $f,g$ approach infinity as well, but it's not through the above equality.)



              (A minor note: a lot of similar laws, invoking for example the sum and difference of functions, also rely on this premise.)






              share|cite|improve this answer














              The limit of the product of functions equals the product of the limits of the functions only if both of those latter limits converge to some finite number. That is to say,



              $$lim_{x to c} f(x)g(x) = lim_{x to c} f(x) cdot lim_{x to c} g(x)$$



              only if $lim_{x to c} f(x)$ and $lim_{x to c} g(x)$ are some finite number. (It is this existence of the limits part that is essential. There are ways to handle cases where $f,g$ approach infinity as well, but it's not through the above equality.)



              (A minor note: a lot of similar laws, invoking for example the sum and difference of functions, also rely on this premise.)







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 26 at 6:42

























              answered Nov 26 at 6:32









              Eevee Trainer

              2,629221




              2,629221












              • Fair point, I think I more or less corrected that nuance.
                – Eevee Trainer
                Nov 26 at 6:44










              • limit [(1 - sinx)/(pi - 2x)^4)(cosx)(8x^3 - pi^3)] as x tends to pi/2. Here the limit of (1-sinx)/(pi-2x)^4 is some non zero finite number. And the limit of rest part of function is 0. Then why the limit of the whole function is not zero?
                – Mathsaddict
                Nov 26 at 7:05


















              • Fair point, I think I more or less corrected that nuance.
                – Eevee Trainer
                Nov 26 at 6:44










              • limit [(1 - sinx)/(pi - 2x)^4)(cosx)(8x^3 - pi^3)] as x tends to pi/2. Here the limit of (1-sinx)/(pi-2x)^4 is some non zero finite number. And the limit of rest part of function is 0. Then why the limit of the whole function is not zero?
                – Mathsaddict
                Nov 26 at 7:05
















              Fair point, I think I more or less corrected that nuance.
              – Eevee Trainer
              Nov 26 at 6:44




              Fair point, I think I more or less corrected that nuance.
              – Eevee Trainer
              Nov 26 at 6:44












              limit [(1 - sinx)/(pi - 2x)^4)(cosx)(8x^3 - pi^3)] as x tends to pi/2. Here the limit of (1-sinx)/(pi-2x)^4 is some non zero finite number. And the limit of rest part of function is 0. Then why the limit of the whole function is not zero?
              – Mathsaddict
              Nov 26 at 7:05




              limit [(1 - sinx)/(pi - 2x)^4)(cosx)(8x^3 - pi^3)] as x tends to pi/2. Here the limit of (1-sinx)/(pi-2x)^4 is some non zero finite number. And the limit of rest part of function is 0. Then why the limit of the whole function is not zero?
              – Mathsaddict
              Nov 26 at 7:05










              up vote
              0
              down vote













              The fact that you decomposed your function into a product with one undetermined factor does not question the truth of the statement all the other answers and comment point out. Just an unfortunate choice of writing your function. Let for instance the inverse function
              $$frac 1 x$$



              It converges, but if you write it as
              $$frac{e^{ix}}{x} e^{-ix}$$



              then the first factor converges to zero and the second one does not converge. The problem is not the algebras on limits (it applies whenever both limits do exist).






              share|cite|improve this answer

























                up vote
                0
                down vote













                The fact that you decomposed your function into a product with one undetermined factor does not question the truth of the statement all the other answers and comment point out. Just an unfortunate choice of writing your function. Let for instance the inverse function
                $$frac 1 x$$



                It converges, but if you write it as
                $$frac{e^{ix}}{x} e^{-ix}$$



                then the first factor converges to zero and the second one does not converge. The problem is not the algebras on limits (it applies whenever both limits do exist).






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  The fact that you decomposed your function into a product with one undetermined factor does not question the truth of the statement all the other answers and comment point out. Just an unfortunate choice of writing your function. Let for instance the inverse function
                  $$frac 1 x$$



                  It converges, but if you write it as
                  $$frac{e^{ix}}{x} e^{-ix}$$



                  then the first factor converges to zero and the second one does not converge. The problem is not the algebras on limits (it applies whenever both limits do exist).






                  share|cite|improve this answer












                  The fact that you decomposed your function into a product with one undetermined factor does not question the truth of the statement all the other answers and comment point out. Just an unfortunate choice of writing your function. Let for instance the inverse function
                  $$frac 1 x$$



                  It converges, but if you write it as
                  $$frac{e^{ix}}{x} e^{-ix}$$



                  then the first factor converges to zero and the second one does not converge. The problem is not the algebras on limits (it applies whenever both limits do exist).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 26 at 6:38









                  Desiderius Severus

                  909618




                  909618






























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