# Math Help - limit (3xy^3)/(x^2+y^2) DNE but why?

1. ## limit (3xy^3)/(x^2+y^2) DNE but why?

I know that the following limit doesn't exist and there's some trick to proving it doesn't exist, anyone know what it is? I think you have to do some substitiution or something. lim x&y -> 0 $(3*x*y^3)/(x^2+y^6)$

I tried showing that the bottom would go to 0 and you can't div by 0, but I got 0 points for the problem with no explanation from the teacher.

2. Originally Posted by bakachan
I know that the following limit doesn't exist and there's some trick to proving it doesn't exist, anyone know what it is? I think you have to do some substitiution or something. lim x&y -> 0 $(3*x*y^3)/(x^2+y^6)$

I tried showing that the bottom would go to 0 and you can't div by 0, but I got 0 points for the problem with no explanation from the teacher.

$\lim_{x,y \rightarrow 0} \dfrac{3xy^3}{x^2 + y^6}$

the limit will exist if all the directional limit exist and are all equal.. so all you ahve to do is to find directions which would result to different limits..

1st.. take $x=y^3$, then

$\lim_{y \rightarrow 0} \dfrac{3y^3y^3}{(y^3)^2 + y^6} = \lim_{y \rightarrow 0} \dfrac{3y^6}{2y^6} = \dfrac{3}{2}$

2nd.. take $x=y$ , then

$\lim_{x,y \rightarrow 0} \dfrac{3xy^3}{x^2 + y^6} = \lim_{y \rightarrow 0} \dfrac{3yy^3}{y^2 + y^6} = \lim_{y \rightarrow 0} \dfrac{3y^4}{y^2(1 + y^4)} = \lim_{y \rightarrow 0} \dfrac{3y^2}{1 + y^4} = 0$

since you found two directions with different limits, therefore the limit does not exist..