Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does the following limit exist?

$$\lim_{(x,y) \to (0,0)}\frac{x^2y^3}{x^4+(x^2+y^3)^2}$$

I tried to solve this problem using polar coordinates, but I can't simplify it. I tried the squeeze theorem, I got $0.5$, but I think this is incorrect.

share|improve this question

2 Answers 2

Let $x^2=y^3$ therefore we have $$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y^3}{x^4+(x^2+y^3)^2}=\lim\limits_{x\to0}\frac{x^4}{5x^4}=\frac{1}{5} \ \ \ (1)$$ and if we consider $x=y$ then $$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y^3}{x^4+(x^2+y^3)^2}=0 \ \ \ \ (2)$$ so that from (1) and (2) we see that limit doesn't exist.

share|improve this answer

Hint: Consider the family of paths $y = kx^{\frac{2}{3}}$.

share|improve this answer
    
Oh i didnt know that we can use to a power of a fraction. –  ys wong Aug 17 at 15:21
    
What if the limit actually exist? –  ys wong Aug 17 at 15:22
    
If the limit exists, the limit along any path exists and the limits along any two paths is the same. Note that using paths you can only show that a two-variable limit does not exist, you can't use paths to prove that a two-variable limit does exist. –  Michael Albanese Aug 17 at 15:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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