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$(x,y) \in \mathbb R^2$

$$\lim_{(x,y)\to(1,1)} \frac{(x-y)^{(x-y)}} {(x-y)}$$

Does the limit above exist? Neither I could compute it nor I could find directions which have different limit values. Can someone help me please? If this limit exists how can I compute it if doesn't exist which directions should I use?

Thanks a lot in advance

user519955
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  • You need to consider paths within the domain of definition for the function that is $$D={(x,y)\in\mathbb{R^2}:x-y>0}$$ – user Dec 07 '18 at 12:32
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    Some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$. Check what is the definition in use in your book, please! – Jyrki Lahtonen Dec 07 '18 at 13:16
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    @JyrkiLahtonen We have already discussed that point recently. Not all sources are to be considered at the same level. Using the first definition is reasonable to a lower (high school) level whereas in a more advanced context the second definition (according to Rudin) should be adopted. Notably according to the first definiiton the interesting question on the present limit would become a trivial question on the domain of the function. – user Dec 07 '18 at 14:32
  • Indeed, @gimusi. I do remember that discussion. I still think that it is important for the askers to be aware of this (and to explain). What you say about the question becoming trivial otherwise is, of course, valid. – Jyrki Lahtonen Dec 07 '18 at 17:28

3 Answers3

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Hint: $\displaystyle\lim_{x\to0}x^x=1$

José Carlos Santos
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This limit does not exist.

Let $\varepsilon \in \mathbb{R}^*_+$.

Take $x=1 + \varepsilon$ and $y= 1$. Then rewrite the quotient as :

$$\varepsilon^{\varepsilon -1} = e^{(\varepsilon -1) \ln (\varepsilon)}.$$ Since $\lim_{\varepsilon \to 0^+} \varepsilon \ln (\varepsilon) =0$ and $\lim_{\varepsilon \to 0^+} - \ln (\varepsilon) = +\infty$, then the limit of $\frac{(x-y)^{(x-y)}} {(x-y)}$ is $+\infty$.

Now you want to apply the same trick with $-\varepsilon$ to get a different limit. The thing is that the power of a negative number might be not be a real number (take for example the square root of $-2$). You can get rid off this problem by using complexe numbers.

Thus, take now $x = 1 - \varepsilon$ and $y=1$. The quotient is now: $$ \frac{(-\varepsilon)^{-\varepsilon}}{-\varepsilon} = e^{-i\frac{\pi}{2}\varepsilon}\frac{\varepsilon^{-\varepsilon}}{-\varepsilon} = e^{-i\frac{\pi}{2}\varepsilon}\frac{1}{-\varepsilon^{\varepsilon + 1}} $$ from here you see that $e^{-i\frac{\pi}{2}\varepsilon}$ tends to 1 and the denominator $-\varepsilon^{\varepsilon + 1} $ goes to $0^{-}$. Thus the limit is now $-\infty$.

Since you can find two "paths" $(x,y)$ which goes to $1$ and which give different limits, the limit $\lim_{(x,y)\to(1,1)} \frac{(x-y)^{(x-y)}} {(x-y)}$ does not exist.

dallonsi
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  • I think that your conclusion is wrong, the limit exists and it is $+\infty$. – user Dec 07 '18 at 12:13
  • Huw do you define $(-\varepsilon)^{-\varepsilon}$? – user Dec 07 '18 at 12:38
  • well if $x,y \in \mathbb{R}$, you can always say that $x^y =|x|^y \times e^{i\arg(x) \times y}$. – dallonsi Dec 07 '18 at 13:17
  • another way to say it is : $\lim_{(x,y) \to (1,1)} (x-y)^{(x-y)} = 1$, right ? So now, if you divide it by $(x-y)$, you can find a path that leads the quotient to $+\infty$ (if $(x-y) \to 0^+$) or $-\infty$ (if $(x-y) \to 0^{-}$). @gimusi, do you agree ? – dallonsi Dec 07 '18 at 13:20
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We have that by $x-y =t \to 0^+$ the given limit reduces to

$$\lim_{(x,y)\to(1,1)} \frac{(x-y)^{(x-y)}} {(x-y)}=\lim_{t\to 0^+} \frac{t^t} {t}\to \infty$$

indeed the points $(x-y)<0$ are not included within the domain of definition for the function that is $$D=\{(x,y)\in\mathbb{R^2}:x-y>0\}$$

Refer also to the related

Edit for a remark

As noticed by Jyrky Lahtonen in the comment "some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$".

That important issue has been deeply discussed here

My observation is that not all sources have to be considered at the same level. Using the first definition is a reasonable way to deal with limits at a lower (high school) level whereas, in a more advanced context, the second more general definition (according to Rudin) should be adopted.

Moreover, note that, according to the first definition, the discussion of limits would reduce, as in the present case, to a (more or less) trivial determination on the domain of definition for functions which is of course a very different topic.

Therefore, when we deal with limits, I strongly suggest to refer to the more general definition excluding points outside the domain of definition for the function considered.

user
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    Why aren't $(x-y) \lt 0$ included in the domain? Domain is all $\mathbb R^2$ – user519955 Dec 07 '18 at 12:17
  • @user519955 The key point is exactly that one. For which values is $f(x)=x^x$ defined? – user Dec 07 '18 at 12:18
  • @user519955 I suggest to refer to that OP and notably to the answer HERE. It often a point of counfusion. Do not hesitate to ask for any clarifiaction on that. – user Dec 07 '18 at 12:24
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    I will glance it. But domain seems like $\mathbb R - {0} \cup {1/(2n | n\in \mathbb N}$ for $x^x$ not $x>0$ to me. For example $(-2)^{(-2)}=1/4$ – user519955 Dec 07 '18 at 12:48
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    @user519955 That’s a good point, I agree that the expression $x^x$ is defined for some values as also for example $(-1/3)^{-1/3}$ but in the calculus context the domain for the function $f(x)=x^x$ is defined only for $x>0$. Refer to that OP for a discussion on that. – user Dec 07 '18 at 13:00