Multivariable Limit with ln(x)











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I need to prove that the following limit is equal to zero:



$$lim limits_{x,y to (1,0)} frac {(x-1)^2ln(x)}{(x-1)^2 + y^2}$$



I tried converting to polar coordinates, since this can sort of simplify the denominator, but wound up stuck with $r cos θ$ inside the logarithmic term. I also tried using the Taylor series for $ln(x)$, but that did not simplify the expression whatsoever.



How do I go about doing this?










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  • Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
    – Robert Z
    Nov 17 at 7:49















up vote
0
down vote

favorite












I need to prove that the following limit is equal to zero:



$$lim limits_{x,y to (1,0)} frac {(x-1)^2ln(x)}{(x-1)^2 + y^2}$$



I tried converting to polar coordinates, since this can sort of simplify the denominator, but wound up stuck with $r cos θ$ inside the logarithmic term. I also tried using the Taylor series for $ln(x)$, but that did not simplify the expression whatsoever.



How do I go about doing this?










share|cite|improve this question
























  • Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
    – Robert Z
    Nov 17 at 7:49













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I need to prove that the following limit is equal to zero:



$$lim limits_{x,y to (1,0)} frac {(x-1)^2ln(x)}{(x-1)^2 + y^2}$$



I tried converting to polar coordinates, since this can sort of simplify the denominator, but wound up stuck with $r cos θ$ inside the logarithmic term. I also tried using the Taylor series for $ln(x)$, but that did not simplify the expression whatsoever.



How do I go about doing this?










share|cite|improve this question















I need to prove that the following limit is equal to zero:



$$lim limits_{x,y to (1,0)} frac {(x-1)^2ln(x)}{(x-1)^2 + y^2}$$



I tried converting to polar coordinates, since this can sort of simplify the denominator, but wound up stuck with $r cos θ$ inside the logarithmic term. I also tried using the Taylor series for $ln(x)$, but that did not simplify the expression whatsoever.



How do I go about doing this?







limits multivariable-calculus






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edited Nov 17 at 7:48









Robert Z

91k1058129




91k1058129










asked Nov 17 at 7:37









Generall Josephina

31




31












  • Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
    – Robert Z
    Nov 17 at 7:49


















  • Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
    – Robert Z
    Nov 17 at 7:49
















Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
– Robert Z
Nov 17 at 7:49




Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
– Robert Z
Nov 17 at 7:49










2 Answers
2






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begin{align}lim limits_{(x,y) to (1,0)} left|frac {(x-1)^2ln(x)}{(x-1)^2 + y^2} right|
&le lim limits_{(x,y) to (1,0)} |ln(x)|\
&=0 end{align}






share|cite|improve this answer




























    up vote
    0
    down vote













    Let




    • $u=x-1$

    • $v=y$


    then



    $$lim_{(x,y) to (1,0)} frac {(x-1)^2 ln(x)}{(x-1)^2 + y^2}=lim_{(u,v) to (0,0)} frac {u^2 ln(1+u)}{u^2 + v^2}=0$$



    indeed



    $$frac {u^2 ln(1+u)}{u^2 + v^2}=frac{ln(1+u)}{u}frac {u^3 }{u^2 + v^2} to 1cdot 0=0$$



    $$frac {u^3 }{u^2 + v^2}=rcdot cos^3 theta to 0$$






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
      2






      active

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      active

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      up vote
      0
      down vote



      accepted










      begin{align}lim limits_{(x,y) to (1,0)} left|frac {(x-1)^2ln(x)}{(x-1)^2 + y^2} right|
      &le lim limits_{(x,y) to (1,0)} |ln(x)|\
      &=0 end{align}






      share|cite|improve this answer

























        up vote
        0
        down vote



        accepted










        begin{align}lim limits_{(x,y) to (1,0)} left|frac {(x-1)^2ln(x)}{(x-1)^2 + y^2} right|
        &le lim limits_{(x,y) to (1,0)} |ln(x)|\
        &=0 end{align}






        share|cite|improve this answer























          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          begin{align}lim limits_{(x,y) to (1,0)} left|frac {(x-1)^2ln(x)}{(x-1)^2 + y^2} right|
          &le lim limits_{(x,y) to (1,0)} |ln(x)|\
          &=0 end{align}






          share|cite|improve this answer












          begin{align}lim limits_{(x,y) to (1,0)} left|frac {(x-1)^2ln(x)}{(x-1)^2 + y^2} right|
          &le lim limits_{(x,y) to (1,0)} |ln(x)|\
          &=0 end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 17 at 7:45









          Siong Thye Goh

          94.8k1462114




          94.8k1462114






















              up vote
              0
              down vote













              Let




              • $u=x-1$

              • $v=y$


              then



              $$lim_{(x,y) to (1,0)} frac {(x-1)^2 ln(x)}{(x-1)^2 + y^2}=lim_{(u,v) to (0,0)} frac {u^2 ln(1+u)}{u^2 + v^2}=0$$



              indeed



              $$frac {u^2 ln(1+u)}{u^2 + v^2}=frac{ln(1+u)}{u}frac {u^3 }{u^2 + v^2} to 1cdot 0=0$$



              $$frac {u^3 }{u^2 + v^2}=rcdot cos^3 theta to 0$$






              share|cite|improve this answer

























                up vote
                0
                down vote













                Let




                • $u=x-1$

                • $v=y$


                then



                $$lim_{(x,y) to (1,0)} frac {(x-1)^2 ln(x)}{(x-1)^2 + y^2}=lim_{(u,v) to (0,0)} frac {u^2 ln(1+u)}{u^2 + v^2}=0$$



                indeed



                $$frac {u^2 ln(1+u)}{u^2 + v^2}=frac{ln(1+u)}{u}frac {u^3 }{u^2 + v^2} to 1cdot 0=0$$



                $$frac {u^3 }{u^2 + v^2}=rcdot cos^3 theta to 0$$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Let




                  • $u=x-1$

                  • $v=y$


                  then



                  $$lim_{(x,y) to (1,0)} frac {(x-1)^2 ln(x)}{(x-1)^2 + y^2}=lim_{(u,v) to (0,0)} frac {u^2 ln(1+u)}{u^2 + v^2}=0$$



                  indeed



                  $$frac {u^2 ln(1+u)}{u^2 + v^2}=frac{ln(1+u)}{u}frac {u^3 }{u^2 + v^2} to 1cdot 0=0$$



                  $$frac {u^3 }{u^2 + v^2}=rcdot cos^3 theta to 0$$






                  share|cite|improve this answer












                  Let




                  • $u=x-1$

                  • $v=y$


                  then



                  $$lim_{(x,y) to (1,0)} frac {(x-1)^2 ln(x)}{(x-1)^2 + y^2}=lim_{(u,v) to (0,0)} frac {u^2 ln(1+u)}{u^2 + v^2}=0$$



                  indeed



                  $$frac {u^2 ln(1+u)}{u^2 + v^2}=frac{ln(1+u)}{u}frac {u^3 }{u^2 + v^2} to 1cdot 0=0$$



                  $$frac {u^3 }{u^2 + v^2}=rcdot cos^3 theta to 0$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 17 at 7:50









                  gimusi

                  88.3k74394




                  88.3k74394






























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