1
$\begingroup$

What is the limit of $$\lim_{(x,y)\to(0,0)} \dfrac {x^4 +y^4}{x^3 +y^3}$$ if it exists?

I have tried to solve it by converting it to polar system $(x,y)=(r\cos a,r\sin a)$ and another settings. However I could not find the limit and not to show that there is no limit.

thanks in advance.

$\endgroup$
3
  • 2
    $\begingroup$ Well, part of the problem is that $x^3+y^3$ can be zero in a neighborhood of $(0,0)$, so you have to be extra careful about your domain. $\endgroup$
    – Thomas Andrews
    Jul 8, 2013 at 13:19
  • 2
    $\begingroup$ you have a problem along the line y= -x. $\endgroup$
    – Lucas Morin
    Jul 8, 2013 at 13:23
  • 3
    $\begingroup$ possible duplicate of Does $\lim_{(x,y) \to (0,0)} \frac{x^4+y^4}{x^3+y^3}$ exist? $\endgroup$
    – user147263
    Sep 7, 2015 at 19:31

2 Answers 2

Reset to default
3
$\begingroup$

The limit does not exist. You can derive multiple limiting values that depend on what path you choose to $(0,0)$. eg. if you choose to follow the path $y=kx$ then you get $(1+k^4)/(1+k^3)$.

Edit: Sorry, right idea, wrong execution (an x got lost somewhere).

Sami's got the same idea: if we head to the origin along $y = -x(1+kx^n)$ we get

$$\lim_{x\to 0} \dfrac{(x^4 + x^4 (1 + kx^n)^4)}{(x^3 - x^3(1 + kx^n)^3)}$$

$$=\lim_{x\to 0} \dfrac{2x^4 + o(x^4)}{-3kx^{n+3} + o(x^{n+3})}$$

$$=\lim_{x\to 0} (-2x^{1-n}/3k)$$

which is $0$ for $n<1$, any positive value we choose for $n=1$, and diverges off to $\infty$ for n>1.

$\endgroup$
2
  • 2
    $\begingroup$ For $y=kx$ I get the limit $lim_{x\rightarrow 0}x\frac{1+k^4}{1+k^3}$... $\endgroup$
    – Avitus
    Jul 8, 2013 at 13:48
  • 1
    $\begingroup$ @Julia Hayward the idea of selecting lines is good: when is $\frac{1+k^4}{1+k^3}$ divergent? Take that $k$ to prove that the original limit... $\endgroup$
    – Avitus
    Jul 8, 2013 at 13:50
3
$\begingroup$

The idea to prove that the limit doesn't exist is to choose a direction such that $x^3+y^3$ is convergent to $0$ faster than $x^4+y^4$ so let $y=-x+x^3$ hence we find $$x^3+y^3=x^3+(-x+x^3)^3=3x^5+o(x^5)\ \text{and}\ x^4+y^4=x^4+(-x+x^3)^4=2x^4+o(x^4)$$ hence $$\lim_{(x,y=-x+x^3)\to(0,0)} \dfrac {x^4 +y^4}{x^3 +y^3}=\lim_{x\to0}\dfrac {2x^4 +o(x^4)}{3x^5 +o(x^5)}=\infty $$

$\endgroup$
1
  • $\begingroup$ That's a fine answer! $\endgroup$
    – amWhy
    May 22, 2014 at 12:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .