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Take the function $f(x,y) = x^y$ where $x \in [0,\infty)$ and $y \in (-\infty,\infty)$.
The value of $f(0,0)$ is indeterminate. I want to know:

Is there some path $S$ on the $(x,y)$-plane that ends at the origin, $(0,0)$, such that $$\lim_{S\to(0,0)} x^y \to\infty$$

And if not, how do you prove that there isn't one?


It's trivial to show that there are limits that can equal $0$ or $1$:

For $x = 0$, $\lim\limits_{y \to 0} \ 0^y = 0 $, and for $y = 0$, $\lim\limits_{x \to 0} \ x^0 = 1 $.

However, I suspect there could be a limit approaching $\infty$ because,

If you take some $\delta$ < 0 that may be arbitrarily close to $0$, then

$$ \lim_{x \to 0} x^{\delta} \to \infty$$

Therefore, in all neighbourhoods centred at $(0,0)$ there exist points at which $x^y$ is arbitrarily large (or approaching infinity if that can be claimed of a point).

It seems, to me, that if there's a path for which $x \to 0$ faster than $y \to 0$ than the value of $x^y$ could keep rising as the path goes on and the limit approaching the origin will approach infinity.

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1 Answer 1

up vote 1 down vote accepted

As you say, a small positive $x$ raised to a very small negative $y$ will be a large positive number. How small must $y$ be? If we want $x^{y}=l$, we simply set $y=\ln(l)/{\ln(x)}$.

Now, define parametric functions $x$ and $y$ by $x(t)=\frac{1}{t}$ and $y(t)=\ln(\ln(t))/\ln(x(t))=\ln(\ln(t))/\ln\left(\frac{1}{t}\right)$. Then the path $$ S(t)=\langle x(t),y(t)\rangle =\left\langle \frac{1}{t},\frac{\ln(\ln(t))}{\ln\left(1/t\right)}\right\rangle$$ has the property that it approaches $\langle 0,0\rangle$ as $t\to\infty$, but the value of $f\circ S(t)$ reduces to $\ln(t)$ which clearly increases without bound.

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So, any path $S(t) =\left\langle \frac{1}{t},\frac{\ln(g(t))}{\ln\left(1/t\right)}\right\rangle$ where $g(t)$ is some increasing function such that as $t \to \infty \implies g(t) \to \infty$ works, right? Does that capture the entire family of solutions? –  Nikhil Mahajan Mar 2 at 6:21
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I think so, except for two caveats. Firstly, you only need $\lim_{t\to\infty}g(t)=\infty$; it doesn't need to be monotone. Secondly, you need to make sure that $g$ doesn't increase too fast, since you need $$y(t)=\frac{\ln(g(t))}{\ln(1/t)}$$ to converge to $0$. –  Unwisdom Mar 2 at 6:25
    
Thanks! That was a great answer. Elegantly solved (in my opinion). –  Nikhil Mahajan Mar 2 at 6:32
    
I just realized, can you not achieve any arbitrary limit $a$ by setting $g(t) = a - e^{-t}$? That seems a little odd. –  Nikhil Mahajan Mar 3 at 14:17

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