Strategy to compute limit of a complex function

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Does a "standard" procedure to compute limits of the complex functions exist?
I know that this question is generic. I expect a generic reply.



Thank you so much in advance.




limits education




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  • l'Hospital's rule... possibly iterated.
    – David G. Stork
    Jul 29 at 5:22






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    "complex function" is abit confusing, better use difficult or complicated
    – gimusi
    Jul 29 at 5:27














up vote
-1
down vote

favorite
1












Does a "standard" procedure to compute limits of the complex functions exist?
I know that this question is generic. I expect a generic reply.



Thank you so much in advance.




limits education




share|cite|improve this question



















  • l'Hospital's rule... possibly iterated.
    – David G. Stork
    Jul 29 at 5:22






  • 1




    "complex function" is abit confusing, better use difficult or complicated
    – gimusi
    Jul 29 at 5:27












up vote
-1
down vote

favorite
1









up vote
-1
down vote

favorite
1






1





Does a "standard" procedure to compute limits of the complex functions exist?
I know that this question is generic. I expect a generic reply.



Thank you so much in advance.




limits education




share|cite|improve this question











Does a "standard" procedure to compute limits of the complex functions exist?
I know that this question is generic. I expect a generic reply.



Thank you so much in advance.





limits education





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 29 at 5:19









Gennaro Arguzzi

301312




301312











  • l'Hospital's rule... possibly iterated.
    – David G. Stork
    Jul 29 at 5:22






  • 1




    "complex function" is abit confusing, better use difficult or complicated
    – gimusi
    Jul 29 at 5:27
















  • l'Hospital's rule... possibly iterated.
    – David G. Stork
    Jul 29 at 5:22






  • 1




    "complex function" is abit confusing, better use difficult or complicated
    – gimusi
    Jul 29 at 5:27















l'Hospital's rule... possibly iterated.
– David G. Stork
Jul 29 at 5:22




l'Hospital's rule... possibly iterated.
– David G. Stork
Jul 29 at 5:22




1




1




"complex function" is abit confusing, better use difficult or complicated
– gimusi
Jul 29 at 5:27




"complex function" is abit confusing, better use difficult or complicated
– gimusi
Jul 29 at 5:27










1 Answer
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When we are dealing with indeterminate form of complicated expression, the standard and more effective method to calculate the limit is of course Taylor's series expansion.



In that way we can reduce any term in polynomial form and the limit becomes easy do solve. The difficult is to select the correct (i.e. minimum) order of expansion and to deal with the remainder terms (usually in little-o or big-O notation).



For example by Taylor's series, with some practice, it is easy to see that



$$lim_xto 0 frac sin^2 left( arctan (log (1+sin x) right) -1+cos(tan x) arcsin^4(sin(log (1-tan(x^2))))+e^sin^2x-1=frac12$$



indeed it correspons to



$$lim_xto 0 frac x^2-1+frac12x^2+o(x^2) x^8+1+x^2-1+o(x^2)=
lim_xto 0 frac frac12x^2+o(x^2) x^2+o(x^2)=frac12$$






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    1 Answer
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    up vote
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    When we are dealing with indeterminate form of complicated expression, the standard and more effective method to calculate the limit is of course Taylor's series expansion.



    In that way we can reduce any term in polynomial form and the limit becomes easy do solve. The difficult is to select the correct (i.e. minimum) order of expansion and to deal with the remainder terms (usually in little-o or big-O notation).



    For example by Taylor's series, with some practice, it is easy to see that



    $$lim_xto 0 frac sin^2 left( arctan (log (1+sin x) right) -1+cos(tan x) arcsin^4(sin(log (1-tan(x^2))))+e^sin^2x-1=frac12$$



    indeed it correspons to



    $$lim_xto 0 frac x^2-1+frac12x^2+o(x^2) x^8+1+x^2-1+o(x^2)=
    lim_xto 0 frac frac12x^2+o(x^2) x^2+o(x^2)=frac12$$






    share|cite|improve this answer



























      up vote
      -1
      down vote













      When we are dealing with indeterminate form of complicated expression, the standard and more effective method to calculate the limit is of course Taylor's series expansion.



      In that way we can reduce any term in polynomial form and the limit becomes easy do solve. The difficult is to select the correct (i.e. minimum) order of expansion and to deal with the remainder terms (usually in little-o or big-O notation).



      For example by Taylor's series, with some practice, it is easy to see that



      $$lim_xto 0 frac sin^2 left( arctan (log (1+sin x) right) -1+cos(tan x) arcsin^4(sin(log (1-tan(x^2))))+e^sin^2x-1=frac12$$



      indeed it correspons to



      $$lim_xto 0 frac x^2-1+frac12x^2+o(x^2) x^8+1+x^2-1+o(x^2)=
      lim_xto 0 frac frac12x^2+o(x^2) x^2+o(x^2)=frac12$$






      share|cite|improve this answer

























        up vote
        -1
        down vote










        up vote
        -1
        down vote









        When we are dealing with indeterminate form of complicated expression, the standard and more effective method to calculate the limit is of course Taylor's series expansion.



        In that way we can reduce any term in polynomial form and the limit becomes easy do solve. The difficult is to select the correct (i.e. minimum) order of expansion and to deal with the remainder terms (usually in little-o or big-O notation).



        For example by Taylor's series, with some practice, it is easy to see that



        $$lim_xto 0 frac sin^2 left( arctan (log (1+sin x) right) -1+cos(tan x) arcsin^4(sin(log (1-tan(x^2))))+e^sin^2x-1=frac12$$



        indeed it correspons to



        $$lim_xto 0 frac x^2-1+frac12x^2+o(x^2) x^8+1+x^2-1+o(x^2)=
        lim_xto 0 frac frac12x^2+o(x^2) x^2+o(x^2)=frac12$$






        share|cite|improve this answer















        When we are dealing with indeterminate form of complicated expression, the standard and more effective method to calculate the limit is of course Taylor's series expansion.



        In that way we can reduce any term in polynomial form and the limit becomes easy do solve. The difficult is to select the correct (i.e. minimum) order of expansion and to deal with the remainder terms (usually in little-o or big-O notation).



        For example by Taylor's series, with some practice, it is easy to see that



        $$lim_xto 0 frac sin^2 left( arctan (log (1+sin x) right) -1+cos(tan x) arcsin^4(sin(log (1-tan(x^2))))+e^sin^2x-1=frac12$$



        indeed it correspons to



        $$lim_xto 0 frac x^2-1+frac12x^2+o(x^2) x^8+1+x^2-1+o(x^2)=
        lim_xto 0 frac frac12x^2+o(x^2) x^2+o(x^2)=frac12$$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 30 at 13:52


























        answered Jul 29 at 5:27









        gimusi

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