Does this limit exist on $\mathbb R^2$

Question

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

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

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

Answer 1

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

Answer 2

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.

Answer 3

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

• domain of $x^x$

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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

• What is $\lim_{x \to 0}\frac{\sin(\frac 1x)}{\sin (\frac 1 x)}$ ? Does it exist?

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.