$ \lim_{(x,y)\rightarrow (0,0)} \frac{x^{5}y^{3}}{x^{6}+y^{4}}. $ Does it exist or not?

Question

I have this limit: $$ \lim_{(x,y)\rightarrow (0,0)} \frac{x^{5}y^{3}}{x^{6}+y^{4}}. $$ I think that this limit does not exist (and wolfram|alpha agrees with me). But I can't find a way to prove it. I chose some paths and the limit was always equal to zero. I tried polar coordinates but I couldn't get an answer. Any tips?

(I also have this one: $$ \lim_{(x,y)\rightarrow (0,0)} x^{y^{2}} $$ Can I get 2 different paths to show that the limit does not exist? $$x=0: \lim_{(x,y)\rightarrow (0,0)} 0^{y^{2}} = \lim_{(x,y)\rightarrow (0,0)} 0 = 0, $$ $$y=0: \lim_{(x,y)\rightarrow (0,0)} x^{0^{2}} = \lim_{(x,y)\rightarrow (0,0)} 1= 1. $$ Am I right? )

Answer 1

The first limit actually exists and is zero. You can show that by definition. Namely, $$\left(\frac{x^5y^3}{x^6+y^4}\right)^2 = \frac{|x^5 y^3|}{x^6+y^4} \cdot \frac{|x^5y^3|}{x^6+y^4}$$

Because $$x^6+y^4\ge x^6 \implies \frac{1}{x^6+y^4} \le \frac{1}{x^6}$$ $$x^6+y^4\ge y^4 \implies \frac{1}{x^6+y^4} \le \frac{1}{y^4}$$ assuming $x$ and $y$ are both non-zero, we have that $$\left(\frac{x^5y^3}{x^6+y^4}\right)^2 \le \frac{|x^5 y^3|}{x^6} \cdot \frac{|x^5y^3|}{y^4} = x^4y^2$$ Taking the square root of both sides we get $$\left|\frac{x^5y^3}{x^6+y^4}\right| \le x^2|y|$$ You can see that this inequality also holds when one of $x$, $y$ is zero (both cannot be zero at the same time). For this to be smaller than $\epsilon$ it suffices to take $\delta=\sqrt[3] \epsilon$.

The second limit does not exist and your proof is correct.

Answer 2

a quick trick worth trying : $$ (x^3 - y^2)^2 \geq 0, $$ $$ x^6 + y^4 \geq 2 x^3 y^2 . $$ $$ (x^3 + y^2)^2 \geq 0, $$ $$ x^6 + y^4 \geq -2 x^3 y^2 . $$ $$ \color{purple}{ x^6 + y^4 \geq 2 |x|^3 y^2 } . $$ when $x,y$ are not both zero, $$ \frac{1}{2} \geq \frac{ |x|^3 y^2}{x^6 + y^4} . $$ $$ \frac{x^2 |y|}{2} \geq \frac{ |x^5 y^3|}{x^6 + y^4} . $$