Limit $\lim_{(x, y) \to (0, \infty)} (xy)$

Question

$$\lim_{(x,y) \to (0, \infty)} (xy) = [x\to\frac{1}{x}\Rightarrow x\to \infty] = \lim_{(x,y)\to(\infty, \infty)} (\frac{y}{x}) = [x = r\cos\theta, y = r\sin\theta] = \lim_{r\to\infty}\frac{r\sin\theta}{r\cos\theta} =\lim_{r\to\infty}\tan\theta$$

Therefore, limit does not exist. Is substitution in the beginning viable here?

Answer 1

More simply we have that as $t\to \infty$

• $x=\frac1t\to 0 \quad y=t\to \infty$

$$xy=\frac1t \cdot t \to 1$$

• $x=\frac1t\to 0 \quad y=t^2\to \infty$

$$xy=\frac1t \cdot t^2=t \to \infty$$

and therefore the limit doesn't exist.

Answer 2

The best way of saying things is this : $\lim_{(x,y) \to (0,\infty)} xy$ exists and equals $L$, if and only if for every subsequence $x_n \to 0$ and $y_n \to \infty$ we have $x_ny_n \to L$, where the latter is convergence of a sequence, in which case we have the $\epsilon-N$ definition of convergence.

Of course, one sees that taking $x_n = \frac 1n$ and $y_n = kn$ for any $k \in \mathbb R_{> 0}$, we have $x_n \to 0$, $y_n \to \infty$ and $x_ny_n \to k$. Therefore, the limit in question does not exist, since different choices of subsequences give different limit values.

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As to what you have done, unfortunately as $x \to 0$ it is not true that $\frac 1x$ goes to $\infty$, since $x$ can approach $0$ from below, in which case $\frac 1x$ assumes negative values and cannot converge to any positive value, let alone positive infinity.

Furthermore, the part where you set $(x,y) = (r \cos \theta,r \sin \theta)$ : The point is that change of variable in limits only works in certain situations. In particular, since the variable $\theta$ has no particular limit as $(x,y) \to (0,\infty)$, I do not think that this change of variable is correct.