Determining the limit











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Find the limit of $$exp(frac{|x-2y|}{(x-2y)^2})$$ 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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    hint $(x-2y)^2=|x-2y|^2$
    – dmtri
    2 hours ago






  • 1




    $e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
    – dmtri
    2 hours ago















up vote
2
down vote

favorite
1












Find the limit of $$exp(frac{|x-2y|}{(x-2y)^2})$$ 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




















  • 1




    hint $(x-2y)^2=|x-2y|^2$
    – dmtri
    2 hours ago






  • 1




    $e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
    – dmtri
    2 hours ago













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





Find the limit of $$exp(frac{|x-2y|}{(x-2y)^2})$$ 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















Find the limit of $$exp(frac{|x-2y|}{(x-2y)^2})$$ 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






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edited 1 hour ago









tonychow0929

24324




24324










asked 2 hours ago









Kashmira

373




373








  • 1




    hint $(x-2y)^2=|x-2y|^2$
    – dmtri
    2 hours ago






  • 1




    $e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
    – dmtri
    2 hours ago














  • 1




    hint $(x-2y)^2=|x-2y|^2$
    – dmtri
    2 hours ago






  • 1




    $e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
    – dmtri
    2 hours ago








1




1




hint $(x-2y)^2=|x-2y|^2$
– dmtri
2 hours ago




hint $(x-2y)^2=|x-2y|^2$
– dmtri
2 hours ago




1




1




$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
– dmtri
2 hours ago




$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
– dmtri
2 hours ago










2 Answers
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It is :



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






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    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






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      2 Answers
      2






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      up vote
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      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

























        up vote
        4
        down vote













        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























          up vote
          4
          down vote










          up vote
          4
          down vote









          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












          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 1 hour ago









          Rebellos

          13.8k31243




          13.8k31243






















              up vote
              2
              down vote













              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

























                up vote
                2
                down vote













                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























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  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












                  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 38 mins ago









                  gimusi

                  92.3k84495




                  92.3k84495






























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