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? )

  • $\begingroup$ Please ask only one question per post. For the first one, the limit does exist; WolframAlpha does a notoriously poor job of dealing with multivariable limits. $\endgroup$ – user296602 Mar 1 '19 at 1:39
  • $\begingroup$ @T.Bongers And how may I prove it? I am so sorry for posting 2 questions, but is it correct? $\endgroup$ – Dr.Mathematics Mar 1 '19 at 1:52
  • $\begingroup$ Has your question been answered? If yes, you should accept an answer. $\endgroup$ – Haris Gušić Jun 24 '19 at 14:38
  • $\begingroup$ For the first question, see math.stackexchange.com/questions/66226/… $\endgroup$ – Arnaud D. Dec 3 '19 at 17:57

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.


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, $$ \frac{1}{2} \geq \frac{ |x|^3 y^2}{x^6 + y^4} . $$ $$ \frac{x^2 |y|}{2} \geq \frac{ |x^5 y^3|}{x^6 + y^4} . $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.