Determining $lim_{(x, y) to (2y, y)} exp(frac{|x-2y|}{(x-2y)^2})$












3












$begingroup$


Find the limit of $$expleft(frac{|x-2y|}{(x-2y)^2}right)$$ when $(x,y) to (2y,y)$.



I have considered two cases: $(x-2y)<0 $ and $(x-2y)>0$.
But in first case the limit turns out to be $0$ and in the second case limit is undefined. I am not sure if my solution is correct or not.






calculus







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$endgroup$








  • 1




    $begingroup$
    hint $(x-2y)^2=|x-2y|^2$
    $endgroup$
    – dmtri
    Dec 15 '18 at 8:36






  • 1




    $begingroup$
    $e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
    $endgroup$
    – dmtri
    Dec 15 '18 at 8:39
















3












$begingroup$


Find the limit of $$expleft(frac{|x-2y|}{(x-2y)^2}right)$$ when $(x,y) to (2y,y)$.



I have considered two cases: $(x-2y)<0 $ and $(x-2y)>0$.
But in first case the limit turns out to be $0$ and in the second case limit is undefined. I am not sure if my solution is correct or not.






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    hint $(x-2y)^2=|x-2y|^2$
    $endgroup$
    – dmtri
    Dec 15 '18 at 8:36






  • 1




    $begingroup$
    $e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
    $endgroup$
    – dmtri
    Dec 15 '18 at 8:39














3












3








3


1



$begingroup$


Find the limit of $$expleft(frac{|x-2y|}{(x-2y)^2}right)$$ when $(x,y) to (2y,y)$.



I have considered two cases: $(x-2y)<0 $ and $(x-2y)>0$.
But in first case the limit turns out to be $0$ and in the second case limit is undefined. I am not sure if my solution is correct or not.






calculus







share|cite|improve this question











$endgroup$




Find the limit of $$expleft(frac{|x-2y|}{(x-2y)^2}right)$$ when $(x,y) to (2y,y)$.



I have considered two cases: $(x-2y)<0 $ and $(x-2y)>0$.
But in first case the limit turns out to be $0$ and in the second case limit is undefined. I am not sure if my solution is correct 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 Dec 15 '18 at 14:27









Namaste

1




1










asked Dec 15 '18 at 8:24









KashmiraKashmira

463




463








  • 1




    $begingroup$
    hint $(x-2y)^2=|x-2y|^2$
    $endgroup$
    – dmtri
    Dec 15 '18 at 8:36






  • 1




    $begingroup$
    $e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
    $endgroup$
    – dmtri
    Dec 15 '18 at 8:39














  • 1




    $begingroup$
    hint $(x-2y)^2=|x-2y|^2$
    $endgroup$
    – dmtri
    Dec 15 '18 at 8:36






  • 1




    $begingroup$
    $e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
    $endgroup$
    – dmtri
    Dec 15 '18 at 8:39








1




1




$begingroup$
hint $(x-2y)^2=|x-2y|^2$
$endgroup$
– dmtri
Dec 15 '18 at 8:36




$begingroup$
hint $(x-2y)^2=|x-2y|^2$
$endgroup$
– dmtri
Dec 15 '18 at 8:36




1




1




$begingroup$
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
$endgroup$
– dmtri
Dec 15 '18 at 8:39




$begingroup$
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
$endgroup$
– dmtri
Dec 15 '18 at 8:39










2 Answers
2






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5












$begingroup$

It is :



$$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
$$=$$
$$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    We have that by $t=x-2y to 0$ we reduce to the simpler



    $$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      5












      $begingroup$

      It is :



      $$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
      $$=$$
      $$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        It is :



        $$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
        $$=$$
        $$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          It is :



          $$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
          $$=$$
          $$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$






          share|cite|improve this answer









          $endgroup$



          It is :



          $$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
          $$=$$
          $$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 8:53









          RebellosRebellos

          15.1k31250




          15.1k31250























              4












              $begingroup$

              We have that by $t=x-2y to 0$ we reduce to the simpler



              $$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                We have that by $t=x-2y to 0$ we reduce to the simpler



                $$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  We have that by $t=x-2y to 0$ we reduce to the simpler



                  $$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$






                  share|cite|improve this answer









                  $endgroup$



                  We have that by $t=x-2y to 0$ we reduce to the simpler



                  $$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 10:05









                  gimusigimusi

                  92.9k84494




                  92.9k84494






























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