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

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

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

• 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. – A.Γ. Dec 10 '18 at 16:29
• Change $y$ to $1/y$. – Paracosmiste Dec 10 '18 at 16:29
• You're assuming $x\to 0^+$. – Shaun Dec 10 '18 at 16:30
• This limit doesn't exist or $0\cdot\infty$ would not be an indeterminate form. – egreg Dec 10 '18 at 17:05

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.

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.