$$\lim_{(x,y)\to (0,0)}\frac{x^2y}{x^3+y^3}$$ Hi, I would like to ask why its limit doesn't exist? Thank you.
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1What do you know about multivariable limits? What sorts of methods do you know for showing they do not exist? – kccu May 14 '17 at 13:48
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If limits exist, they are unique. Can you evaluate this limit in two ways such that they do not yield the same answer? – Meneer-Beer May 14 '17 at 13:53
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Possible duplicate of Determining existence of limit with multiple variables: $\lim_{(x,y)\to (0,0)} \frac{xy^2}{x^3+y^3}$ – Martin Sleziak May 14 '17 at 16:29
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Consider the limits on two ways:
1) First fix $x=0$ and consider the limit as $y$ approaches 0.
2) Along the path $y=x$.
Soby
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Hint:
Let $(x,y)\to(0,0)$ on the $y$ axis, observe if there is a limit on this path.
Then, let $(x,y)=(t,t),$ as $t\to 0$, observe if there is a limit, if there is one compare it with the former.
Ángel Mario Gallegos
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Along the line $y=x $, the limit is $$\frac {1}{2} $$ and along $y=2x $, the limit is $$\frac{2}{9} $$ thus it doesn't exist.
hamam_Abdallah
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To show that this limit does not exist, we need to show that there exists two paths to the point $(0,0)$ that approach different values. 1) Consider the path along $y = x$. Now $$\lim_{(x,y)\rightarrow (0,0)} \frac{x^2y}{x^3+y^3} = \lim_{(x,y)\rightarrow (0,0)} \frac{x^2(x)}{x^3+(x)^3} = \lim_{(x,y)\rightarrow (0,0)} \frac{x^3}{2x^3} = \lim_{(x,y)\rightarrow (0,0)} \frac{1}{2} = \frac{1}{2}\cdot$$ 2) Consider the path along $y = 2x$. Now $$\lim_{(x,y)\rightarrow (0,0)} \frac{x^2y}{x^3+y^3} = \lim_{(x,y)\rightarrow (0,0)} \frac{x^2(2x)}{x^3+(2x)^3} = \lim_{(x,y)\rightarrow (0,0)} \frac{2x^3}{9x^3} = \lim_{(x,y)\rightarrow (0,0)} \frac{2}{9} = \frac{2}{9}\cdot$$ Therefore, the $\lim_{(x,y)\rightarrow (0,0)} \frac{x^2y}{x^3+y^3}$ does not exist.
Ahmed
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