Difficult limit of two variables

Question

Find $$\lim_{(x,y)\to (0,0)}{x^2y^3 \over x^5+y^3}$$

I thought that this limit doesn´t exists so I tried every possible path of the form $x^\alpha$ but the result is $0$, so maybe the limit exists but I don´t know how to prove it. Any idea or suggestion would be highly appreciated.

Answer 1

The quotient is not even defined whenever $x^5+y^3 = 0$, which happens for $(x,y) = (t^3, -t^5)$ with $t \in \mathbb{R}$. So it is sort of irrelevant to take limits, except if you implicitly exclude approaching $(0,0)$ along this set.

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As mentioned in the comments, it is implicitly assumed that we exclude the domain where the fraction is undefined. In order to merit the upvotes this answer received, I shall add a few comments on how one could think about the problem and incidentally be led to the solutions given in the other answers.

The fraction is undefined on the set $S = \{ (x,y) \in \mathbb{R}^2 | x^5+y^3=0\} = \{ (t, -t^{5/3}) \in \mathbb{R}^2 | t \in \mathbb{R}\}$. Observe that the numerator of the fraction is nonzero on $S \setminus \{(0,0)\}$, so that the absolute value of the fraction (whenever defined) is quite large in any sufficiently small neighborhood of any point $(x_0, y_0) \in S \setminus \{(0,0)\}$. This suggests that if one approches $(0,0)$ along a curve inside $\mathbb{R}^2 \setminus S$ which is 'sufficiently close' to the set $S$, then the associated limit should be divergent. What means 'sufficiently close' here depends on the numerator, since $x^2y^5$ tends to $0$ at some appreciable speed as one approaches $(0,0)$ along any path, thus balancing to some extend the divergence induced by the denominator.

It is thus appropriate to look for a curve of the form $t \mapsto (t, -t^{5/3} + t^{\alpha})$ with $\alpha > 0$. The larger $\alpha$ is, the 'closer' the path is to $S$ while still avoiding it (except at $(0,0)$ of course). Making the substitution in the fraction for $t \neq 0$, one gets

$$ \begin{align} \notag \frac{t^2(-t^{5/3} + t^{\alpha})^3}{t^5 + (-t^{5/3} + t^{\alpha})^3} &= \frac{t^2(-1 + t^{\alpha - 5/3})^3}{1 + (-1 + t^{\alpha -5/3})^3} \underset{t \to 0}{\sim} \frac{t^2(-1 + 3t^{\alpha - 5/3})}{1 + (-1 + 3t^{\alpha -5/3})} = \frac{- t^{11/3 - \alpha} + 3t^{2}}{3} \,. \end{align}$$

If one takes $\alpha > 11/3$, for instance $t = 4$, then the leading term as $t \to 0$ is $ (-1/3)t^{11/3 - \alpha}$, which diverges in this limit.

Anticipating this result, one could have taken from the beginning a very large $\alpha$ compared to all the numbers present in the fraction, for instance $\alpha = 10$, thereby avoiding the necessity to find what is a 'sufficiently large $\alpha$'.

Answer 2

Hint: Take the path $$\gamma(t)=\left(t,\frac{t^5}{t^2-1}\right).$$

Answer 3

HINT: Crucial is the behaviour near the path $y^3=-x^5$. Let, e.g., $y^3=-x^5+x^{10}$. Then $$ \lim_{x\to0}\frac{x^2(-x^5+x^{10})}{x^{10}}=\lim_{x\to0}\frac{-x^7+x^{12}}{x^{10}}. $$ It is easy to show that this limit does not exist.

Answer 4

Note that

for $x=t$ and $y=t$ with $t\to0$

$$\lim_{(x,y)\to (0,0)}{x^2y^3 \over x^5+y^3}=\lim_{t\to 0}{t^5 \over t^5+t^3}=\lim_{t\to 0} {t^2 \over t^2+1}=0$$

whereas for $x=t^{\frac15}$ and $y=-(t+t^2)^{\frac13}$ with $t\to0$

$$\lim_{(x,y)\to (0,0)}{x^2y^3 \over x^5+y^3}=\lim_{t\to 0} {-t^{\frac25}(t+t^2) \over t-(t+t^2)}=\lim_{t\to 0} {-t^{\frac25}(t+t^2) \over -t^2}=\lim_{t\to 0} {t^{\frac25}(1+t) \over t}=\\=\lim_{t\to 0} {1+t \over t^{\frac35}}\to \pm\infty$$

then the limit does not exist.