Towering limits

Calculus Level 5

$\Large \lim_{\substack{x \to 0^+ \\ y \to 0^+}}{\frac{x^{y^x}-x^{y^2}}{y^x}}=\,?$

-\infty

-1

0

1

\infty

Limit does not exist

1 solution

Calvin Lin Staff
Jan 15, 2018

All limits are to $0^+$ unless otherwise stated.

Taking the path $(x,x)$ , we get the limit of $\lim \frac{ x^{x^x} - x^{x^2} } { x^x } = \frac{0 - 1}{1} = -1$ .
Here, we use the fact that $\lim x^x = 1$ , $\lim x^{x^2} = 1$ and $\lim x^{x^x} = 0$ , which can be shown by taking logs.

Taking the path $( - \frac{1}{\ln y } , y )$ , we get the limit of $\lim \frac{ x^ { \frac{1}{e} } - x^y } { \frac{1}{e}} = \lim e x^y = e$ .
Here, we use the fact that $y^{ - \frac{1}{ \ln y } } = \frac{1}{e}$ and $\lim ( - \frac{1}{\ln y})^y = 1$ , which can be shown by taking logs.

Hence, the limit (over all possible paths) does not exist.

Consider $\lim x^y$ . We know that $\lim x^x = 1$ , and want to understand why $\lim x^y$ does not exist.

Useful Lemma: If $f(x)$ and $g(x)$ are polynomials such that $x\rightarrow 0^+, f(0), g(0) \rightarrow 0^+$ , then $\lim_{x \rightarrow 0} f(x) ^ { g(x) } = 1$ .
This follows because the linear term is the most important contributor, and we know that $\lim (ax)^{(bx) } = 1$ for (positive) constants $a, b$ .

Most people would be checking (somewhat) polynomial paths, which is why they always get the limiting value of 1. However, we can see that to get a case where $\lim x^y \neq 1$ , we have to avoid polynomial paths. For example, the logarithmic path $(x, - \frac{1}{ \ln x })$ works and we get $\lim x^y = \lim \frac{1}{e} = \frac{1}{e}$ .

In fact, by taking logarithms, we want to check against $\lim g(x) \ln f(x)$ . From here, we can start to deduce how to get an instance where $\lim f(x) ^ { g(x) } \neq \lim g(x) ^ { f(x) }$ .

(Might not be true, but I believe it is) Relating back to this problem. If we kept to polynomial paths, we will always get the limit of -1.

Moderator note:

Remember that for limits, you have to consider all possible paths towards that point.

In 1-variable tending to $0$ , there are essentially two paths - the left-hand path and the right-hand path (along with the path that goes back and forth between the left hand and right-hand path). This is why we say the limit exists if and only if the left-hand and right-hand limit exists.

In 2-variables tending to $(0,0)$ , there are a lot of paths which are essentially different. It is not sufficient to only consider paths of the type $y = mx$ . We also have paths like $y = mx^2 , y = e^x -1$ , and those which spiral around the point.

To show that a 2-variable limit does not exist, it is sufficient to find 2 paths with different limits, or a path which has no limit.

To prove that a 2-variable limit exists, it is almost impossible to consider all possible paths. As such, it is often better to use the delta-epsilon definition of a limit , which states that

Let $f(x,y)$ be a function defined on an open interval around $x_0, y_0$ ( $f(x_0, y_0)$ need not be defined). We say that the limit of $f(x)$ as $(x,y)$ approaches $(x_0, y_0)$ is $L$ , i.e. $\lim _{ (x,y) \to (x_{0},y_0) }{f(x) } = L,$ if for every $\varepsilon > 0$ there exists $\delta >0$ such that for all $x$ $0 < \left| x - x_{0} \right | + \left| y - y_{0} \right | <\delta, \textrm{ } \Rightarrow \textrm{ } \left |f(x) - L \right| < \varepsilon.$

You are saying that $0^{y^2} = 1$ . No it isn't; it is $0$ . Moreover, the part $(0,y)$ is not possible (since $x > 0$ throughout), except after a limiting process.

Is it worth taking the time to show that the existence of the double limit would prove the existence of these directional limits (including the ones along $(x,0)$ and $(0,y)$ ), and that all these limits are equal to the double limit?

Mark Hennings - 3 years, 4 months ago

Ah yes. That will need a fix.

Isn't the second obvious by definition, since the existence of the double limit is taken also every possible path? Also, I would like to stress that it's not just directional limits that we have to care about. We also need polynomial paths and spirals.

Calvin Lin Staff - 3 years, 4 months ago

It is obvious, yes, but questions of limits are done badly, so an example of good practice would be a good idea.

Mark Hennings - 3 years, 4 months ago

@Mark Hennings – Alright, I've made a bunch of edits. Thoughts?

Calvin Lin Staff - 3 years, 4 months ago

@Calvin Lin – The proof is fine, but the subsequent discursion may still be a little off...

If the double limit exists, all path limits will exist and be the same. Is the converse itrue, unless you have some uniformity on the rate of convergence across all paths.

Mark Hennings - 3 years, 4 months ago

Towering limits

1 solution

Moderator note:

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