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

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

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    $\begingroup$ Not really. If $x\to 0^+$ (from one side) then $\frac{1}{x}\to +\infty$, but if $x\to 0$ oscillating around zero, then the limit $1/x$ does not exists. $\endgroup$ – A.Γ. Dec 10 '18 at 16:29
  • $\begingroup$ Change $y$ to $1/y$. $\endgroup$ – Paracosmiste Dec 10 '18 at 16:29
  • $\begingroup$ You're assuming $x\to 0^+$. $\endgroup$ – Shaun Dec 10 '18 at 16:30
  • $\begingroup$ This limit doesn't exist or $0\cdot\infty$ would not be an indeterminate form. $\endgroup$ – egreg Dec 10 '18 at 17:05
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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.

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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.


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.

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