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Why does this limit not exist?

$$\lim\limits_{ (x,y) \to (0,0) } \frac{x-y}{\sqrt x - \sqrt y}$$

If you set y = 0, the limit goes to zero. If you set x = 0, the limit goes to zero. You can also manipulate it with algebra to get zero. However, if x=y you have zero/zero before you even evaluate the limit but is that proof enough?

Thanks!

From Larson Calculus 13.2 Exercise 27

Mike
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  • What is the definition of the limit? You need to feed your observations into that. – Mark Bennet Mar 25 '17 at 06:48
  • Try to use two paths to evaluate the limit. Common paths include the $x$ axis, the $y$ axis and the line $x=y$ – codetalker Mar 25 '17 at 06:50
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    When you say $(x,y)\to (0,0)$ you certainly mean "goes to zero inside the largest set where the expression (function) is defined". For example, negative $x$ and $y$ are obvious troubles as well as $x=y$ and must be excluded. The mere presence of such points does not disprove the existence of the limit by itself. – A.Γ. Mar 25 '17 at 07:23
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    Where is your function defined ? Remember that the definition of a limit depends on the domain of definition of the function! – Thomas Mar 25 '17 at 07:26
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    @Siddhant And so what? This seems to miss the point... – Did Mar 25 '17 at 10:36

2 Answers2

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The main problem with this exercise is the following: The expression $$\Psi(x,y):={x-y\over\sqrt{x}-\sqrt{y}}$$ is undefined when $x<0$ or $y<0$, or $x=y$. In this situation one can argue in two ways:

(i) You can say that as a general rule the domain of definition of an expression is the set of $(x,y)$ for which it can be evaluated without asking supplementary questions. In the case at hand this is the set $$\Omega:=\{(x,y)\in{\mathbb R}^2\>|\>x\geq0,\ y\geq0, \ x\ne y\}\ .$$ The point $(0,0)$ is a limit point of $\Omega$, hence it makes sense to consider $\lim_{(x,y)\to(0,0)}\Psi(x,y)$. Now for all $(x,y)\in\Omega$ one has $$\Psi(x,y)={x-y\over\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\ ,\tag{1}$$ and here the right hand side obviously converges to $0$ when $(x,y)\to(0,0)$.

(ii) You can say that the identity $(1)$ allows to extend the function defined by ${x-y\over\sqrt{x}-\sqrt{y}}$ continuously to all of $\bigl({\mathbb R}_{\geq0}\bigr)^2$. This is like defining ${\sin x\over x}$ to be $1$ at $x=0$. But this is a voluntary act, and is not stipulated in the formulation of the problem. If you want to adopt this position then the limit is of course again $=0$, since the function $(x,y)\mapsto\sqrt{x}+\sqrt{y}$ is continuous at $(0,0)$.

Christian Blatter
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Well it is the same than $\displaystyle{\frac{x^2}{x}}$.

Is it a $\frac 00$ limit or do we use algebra to simplify this to $x$ and say it is going to $0$ ?

In fact you can manipulate any expression to make it undefined if you want. But we make a postulate that for a function $f$ if after algebric manipulations the new function $\tilde f$ is well defined then we can extend $f$ by continuity to $\tilde f$.

In general we do not detail all of these considerations, we just assume the extension with the maximum domain fo definition.

In our case, we have

$\begin{cases} \text{on }A=\{x\ge 0,\ y\ge 0,\ x\neq y\} & f(x,y)=\frac{x-y}{\sqrt x-\sqrt y} \\ \text{on }\bar A=\{x\ge 0,\ y\ge 0\} & \tilde f(x,y)=\sqrt x+\sqrt y \end{cases}$

It is easy to see that :

  • $f$ is continuous on $A$ as a composition of continuous functions
  • $\tilde f$ is well defined and continuous on $\bar A$
  • the restriction to $A$ of $f$ is $\tilde f:\qquad f_{/A}=\tilde f$

So we generally choose to work with $\tilde f$ defined on $\bar A$ instead of $f$ defined on $A$, so as to get rid of the somehow artificial singularity when $x=y$.

Now for the original question : $\tilde f(0,0)=0$ so the limit for $f$ does exists and we can choose to extend $f$ by continuity in this point by setting $f(0,0)=0$.


Edit:

A little more into details about extending by continuity.

The theorem says : if a function $f$ is cauchy-continuous on a domain $A$ then it can be extended in a unique way to a function $\tilde f$ continuous on $\bar A$


For a Cauchy sequence in $A$ we have $f(x_n,y_n)=\tilde f(x_n,y_n)=\sqrt{x_n}+\sqrt{y_n}$ which is still cauchy according to the post forward for $x_n\ge 1$ and $y_n\ge 1$ : $\sqrt{a_n}$ is also cauchy if $a_n\ge 1$ is cauchy

For $[0,1]\times[0,1]$ this is not an issue either since this is a compact on which $\tilde f$ is continuous.

Finally the two remaining bands are out of interest, since $f=\tilde f$ already on these.

Thus $f$ is continuously extentable to $\bar A$ by $\tilde f$.


So it does not matter that it is a single point $x=0$ like in the case $\frac{x^2}{x}$ or a line $x=y$ like in the case of $f(x)$, what's matter is that in the restricted domain of definition the function is Cauchy-continuous (given that $(\mathbb R,|\cdot|)$ is complete, we can always conclude positively that the extension is sound).


Also a Cauchy sequence being bounded (let say inside a compact $K$), what's really important is not the behaviour at infinity but the continuity of the extension near the singularity. We want $\tilde f$ simply continuous over $\bar A\cap K$. ($\tilde f$ is then continuous over a compact, thus uniformly continuous, thus cauchy continuous). It is more constraining than just simple continuity of $f$ over $A$.


For instance the counter example given by Hanul Jeon is the above mentionned post, does not work because $f(x)=1/x,\ a_n=1/n$ then $\tilde f$ would not be continuous in $0$ no matter which value you assign to $\tilde f(0)$.

Community
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zwim
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  • It is not the same as $\dfrac{x^2}{x}$, because that expression is always defined when you're not at $x=0$, and its ordinarily limit is defined unambigously without having to restrict the domain. The concern here is that the expression is not defined for points arbitrarily close but not equal to $(x,y)=(0,0)$, as the later part of this answer deals with. – Jonas Meyer Mar 25 '17 at 06:55
  • @JonasMeyer There are other inconsistencies in the original limit, for example, negative $x$ and $y$. I guess what the OP means is $(x,y)\to(0,0)$ within the definition domain for the expression, which excludes, in particular, $x=y$, so in a way it is, if not "the same", similar to the case $\frac{x^2}{x}$. – A.Γ. Mar 25 '17 at 07:16
  • I agree with most of your comment but for the last part here is why I don't see it that way: For the typical definition of $\lim\limits_{x\to a}f(x)$, definition of $f$ at $a$ is irrelevant. For the typical definition of $\lim\limits_{(x,y)\to(a,b)}f(x,y)$, definition of $f$ at $(a,b)$ is irrelevant but it is expected that $f(x,y)$ is defined for $(x,y)$ close but unequal. As you pointed out even beyond the $y=x$ the OP left out the other domain restriction, and I agree that with the domain restriction all becomes good. – Jonas Meyer Mar 25 '17 at 07:29
  • In 1D, perhaps a closer analogy would be $$\lim\limits_{x\to 0}x\cos(\sec(1/x))$$ because it is easy to show the expression goes to $0$ as $x\to 0$ on the natural domain of the function, and $0$ is a limit point of the domain, but the function is undefined arbitrarily close to $0$. Change the $x$ to $\sqrt x$ for even closer analogy perhaps. – Jonas Meyer Mar 25 '17 at 07:32
  • Or for better arbitariness of nondefinition in 1D, how about $$f(x) = \sqrt x\prod\limits_{n=1}^\infty\frac{x-\frac1n}{x-\frac1n}$$ – Jonas Meyer Mar 25 '17 at 07:38
  • I tried to be more thorough about continuity extension for a function of 2 variables. I hope it is sound, it is easy to be mistaken with these delicate considerations. – zwim Mar 25 '17 at 08:45
  • Thank you for your answer! I am going to read that cauchy continious link and see if I can understand it. – Mike Mar 25 '17 at 16:57