Limit in two variables at infinity

[dunkelheit]

I have a doubt in the resolution of this limit: given the conditions [MATH]x>0[/MATH] and [MATH]y>0[/MATH] evaluate

[MATH]\lim_{x^2+y^2 \to \infty} xy[/MATH]​

The solution says this: "The limit doesn't exist since on the restriction [MATH]f(0,t)[/MATH] the function is [MATH]0[/MATH] and so its limit is [MATH]0[/MATH] but on the restriction [MATH]f\left(t,\frac{1}{t}\right)[/MATH] it is
[MATH]\lim_{t \to \infty} t \frac{1}{t}=1[/MATH]and so the limit doesn't exist."
My doubt is: when the restriction [MATH]f\left(t,\frac{1}{t}\right)[/MATH] is used the limit is considered as [MATH]t\to\infty[/MATH] because, I suppose, the condition [MATH]x^2+y^2 \to \infty[/MATH] must be satisfied; so in this case [MATH]x^2+y^2[/MATH] becomes [MATH]t^2+\frac{1}{t^2}[/MATH] and of course [MATH]t^2+\frac{1}{t^2} \to \infty[/MATH] as [MATH]t \to \infty[/MATH].
But [MATH]t^2+\frac{1}{t^2} \to \infty[/MATH] if [MATH]t \to 0^+[/MATH] too, so the question is: is equivalent to consider the limit as MATH]t\to 0^+[/MATH] instead of $\to \infty$ in this particular case?
More in general, in limits as [MATH]x^2+y^2 \to \infty [/MATH] is correct to let [MATH]t\to t_0[/MATH] if on a restriction [MATH]f(g(t),h(t))[/MATH] it is satisfied [MATH]g^2(t)+h^2(t) \to \infty[/MATH] as [MATH]t \to t_0[/MATH]?

 

[Dr.Peterson]

I have a doubt in the resolution of this limit: given the conditions [MATH]x>0[/MATH] and [MATH]y>0[/MATH] evaluate

[MATH]\lim_{x^2+y^2 \to \infty} xy[/MATH]​

The solution says this: "The limit doesn't exist since on the restriction [MATH]f(0,t)[/MATH] the function is [MATH]0[/MATH] and so its limit is [MATH]0[/MATH] but on the restriction [MATH]f\left(t,\frac{1}{t}\right)[/MATH] it is
[MATH]\lim_{t \to \infty} t \frac{1}{t}=1[/MATH]and so the limit doesn't exist."
My doubt is: when the restriction [MATH]f\left(t,\frac{1}{t}\right)[/MATH] is used the limit is considered as [MATH]t\to\infty[/MATH] because, I suppose, the condition [MATH]x^2+y^2 \to \infty[/MATH] must be satisfied; so in this case [MATH]x^2+y^2[/MATH] becomes [MATH]t^2+\frac{1}{t^2}[/MATH] and of course [MATH]t^2+\frac{1}{t^2} \to \infty[/MATH] as [MATH]t \to \infty[/MATH].
But [MATH]t^2+\frac{1}{t^2} \to \infty[/MATH] if [MATH]t \to 0^+[/MATH] too, so the question is: is equivalent to consider the limit as [MATH]t\to 0^+[/MATH] instead of [MATH]\to \infty[/MATH] in this particular case?
More in general, in limits as [MATH]x^2+y^2 \to \infty [/MATH] is correct to let [MATH]t\to t_0[/MATH] if on a restriction [MATH]f(g(t),h(t))[/MATH] it is satisfied [MATH]g^2(t)+h^2(t) \to \infty[/MATH] as [MATH]t \to t_0[/MATH]?

I suspect that they are defining a "restriction" as a path as t goes to infinity, not the entire parametric curve.

But since [MATH]x^2+y^2[/MATH] goes to infinity in both cases, whether t goes to infinity or to zero, does it matter? Sketch the curve (t, 1/t), and you can see that it goes to infinity in two directions, both within quadrant I. Either case disproves the limit. They just picked one direction to use as their example.