# Determine whether $\lim_{(x,y)\to (2,-2)} \dfrac{\sin(x+y)}{x+y}$ exists.

I am trying to determine whether $\lim_{(x,y)\to (2,-2)} \dfrac{\sin(x+y)}{x+y}$ exists. I should be able to use the following definition for a limit of a function of two variables:

Let $f$ be a function of two variables that is defined on some open disk $B((x_0,y_0),r)$, except possibly at the point $(x_0,y_0)$ itself. Then

$$\lim_{(x,y)\to (x_0,y_0)}f(x,y)=L$$

if for any $\varepsilon>0$, however small, there exists a $\delta>0$ such that

$$\text{if } 0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta \text{ then } |f(x,y)-L|<\varepsilon$$

The limit definition above applied to the limit $\lim_{(x,y)\to (2,-2)} \dfrac{\sin(x+y)}{x+y}$ requires that the function $\dfrac{\sin(x+y)}{x+y}$ be defined on every point of some open disk $B((2,-2),r)$ (with the possible exception of the point $(2,-2)$).

My problem here is that the function $\dfrac{\sin(x+y)}{x+y}$ is undefined for $x+y=0$, that is, it is undefined for all points in the line $y=-x$. If I understood it correctly, it seems to mean that I can't find any open disk $B((2,-2),r)$ on which the given function is defined. Therefore, it seems that I can't apply the limit definition.

What am I misunderstanding here?

The argument above makes me believe that this limit does not exist (because I can't apply the definition), but Wolfram|Alpha says that the value of this limit is $1$. Why this apparent contradiction?

Edit: I should point out that this problem and the limit definition that I'm using are both taken from the same book. So, the book assumes that I will use the given limit definition.

## Update

I have one more doubt concerning this.

The same book that I'm using says that the function below is continuous for all points $(x,y)$ in $R_2$.

$$f(x,y)=\begin{cases} \dfrac{\sin(x+y)}{x+y} & \text{ if x + y \neq 0} \\ 1 & \text{ if x + y = 0}\end{cases}$$

The proof is as follows:

Let $h(t)=\begin{cases} \dfrac{\sin(t)}{t} & \text{ if$t \neq 0$} \\ 1 & \text{ if$t = 0$}\end{cases}$ and $g(x,y) = x+y$; so, $f(x,y) = h(g(x,y))$. The domain of $g$ is $R_2$ and it is continuous everywhere in its domain. Since $g(x,y)\to 0$ as $x+y\to 0$ and $h$ is continuous at $0$, $h(g(x,y))$ is continuous everywhere.

The definition of continuity in this book states that, for a two-variable function $g$ to be continuous at $(x_0,y_0)$, $g(x_0,y_0)$ has to exist and $\lim_{(x,y)\to (x_0,y_0)} g(x,y)$ has to be equal to $g(x_0,y_0)$. So, by the given definition of continuity, $f(x,y)$ being continuous everywhere means that $\lim_{(x,y)\to (x_0,y_0)} f(x,y) = f(x_0,y_0)$ for all $(x_0,y_0)$.

The proof above seems to imply that $\lim_{(x,y)\to (2,-2)} \dfrac{\sin(x+y)}{x+y}$ should exist. But the problem is that this limit should not exist, considering the definition of limit that I am using in this question. Is there a contradiction here or am I missing something?

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You're not misunderstanding at all. :) – Antonio Vargas Jan 21 '13 at 20:41
If it is a problem invented by you, then you may use that definition of limit, and conclude that it does not exist. If it is a problem taken from somewhere else, you should use the definition of limit in that source. – GEdgar Jan 21 '13 at 21:38
@GEdgar: Both the problem and the definition are taken from the same book. – anonymous Jan 21 '13 at 21:42
Then you are right, but Wolfram|Alpha is not that book, since it seems to be using a different definition! – GEdgar Jan 21 '13 at 21:44
I suggest that you email the graph in my answer to the author of the book and ask him if he really thinks that this limit doesn't exist. Note that if you extend the function to be $1$ on the line $x+y=0$ (except precisely at $(2,-2)$), the limit exists even with your definition. – mrf Jan 21 '13 at 21:53

Consider the situation near $(2,-2)$:

$\hspace{4.5cm}$

Notice that between the two black lines, in particular inside the circle, $|x+y|\le0.00001$.

We know that $\lim\limits_{x\to0}\frac{\sin(x)}{x}=1$, so by making the circle small enough, $\frac{\sin(x+y)}{x+y}$ can be made as close to $1$ as we like inside the circle. So off of the line $x+y=0$, $$\lim\limits_{\substack{x\to2\\y\to-2}}\frac{\sin(x+y)}{x+y}=1$$ However, strictly speaking, $\frac{\sin(x+y)}{x+y}$ does not exist on the line $x+y=0$, so the limit does not exist.

Concerning the Update:

As I said above, off the line $x+y=0$, $$\lim\limits_{\substack{x\to2\\y\to-2}}\frac{\sin(x+y)}{x+y}=1$$ The reason the line $x+y=0$ is excluded, is that there are approaches along that line where the values of $\frac{\sin(x+y)}{x+y}$ are not defined because division by $0$ would occur. If we define $$f(x,y)=\left\{\begin{array}{} \frac{\sin(x+y)}{x+y}&\text{if }x+y\ne0\\ 1&\text{if }x+y=0 \end{array}\right.$$ then $f(x,y)$ is defined everywhere and for the same reason that $$g(x)=\left\{\begin{array}{} \frac{\sin(x)}{x}&\text{if }x\ne0\\ 1&\text{if }x=0 \end{array}\right.$$ is continuous, $f(x,y)=g(x+y)$ is continuous. In fact, $f$ is the composition of two continuous functions, $g$ and addition.

However, unless we explicitly define $f(x,y)$ along the line $x+y=0$, $f$ is not even defined on that line, much less continuous there.

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+1. Thank you for the explanation. It seems that the definition of limit that I'm using is more strict than it could be. – anonymous Jan 21 '13 at 22:53
I've added one more doubt to the original question, under the title Update. Thank you in advance. – anonymous Jan 22 '13 at 20:43

The exact defintion of limit varies a bit. In many situations it's too restrictive to assume that f is defined on a punctured open neighboorhood of the point. Often it's enough to only require that the domain of definition intersects every such neighbourhood.

With this more permissive definition, the limit can be calculated using a well-known one-variable limit:

$$\lim_{(x,y)\to(2,-2)} \frac{\sin(x+y)}{x+y} = \lim_{t\to 0} \frac{\sin t}{t} = 1$$

(since the value of $f(x,y) = \frac{\sin(x+y)}{x+y}$ only depends on $x+y$ which clearly tends to $0$ as $(x,y) \to (2,-2)$).

The following graph of $z = \frac{\sin(x+y)}{x+y}$ should make the calculation above even more believable:

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+1 nice answer. However, it is not said that the OP can choose his own definitions for the limit, especially if this is homework. – CBenni Jan 21 '13 at 21:40
@CBenni Agreed, that's why I started with a caveat. Most books only require the inequality in the $\varepsilon$-$\delta$ defintion to hold for points that are actually in the domain of definition. (Not every book makes this explicit, though.) – mrf Jan 21 '13 at 21:43
+1. Thank you for the explanation. – anonymous Jan 21 '13 at 22:52
@mrf: I've added one more doubt to the original question, under the title Update. Thank you in advance. – anonymous Jan 22 '13 at 20:42

Using the bona fide interpretation of the expression $$f(x,y):={\sin(x+y)\over x+y},$$ it is defined on the set $D:={\mathbb R}^2\setminus\{(x,-x)\ |\ x\in{\mathbb R}\}$. The point $(2,-2)$ is an accumulation point of $D$; therefore it is allowed to consider the $\lim_{(x,y)\to(2,-2)} f(x,y)$. Here it is tacitly understood that $f$ is only evaluated in points of $D$.

The function $${\rm sinc}:\quad {\mathbb R}\to{\mathbb R},\qquad u\mapsto\cases{{\sin u\over u}\quad &(u\ne0) \cr \ 1&(u=0)\cr}$$ is well known to be continuous on all of ${\mathbb R}$; therefore the function $$\tilde f(x,y):={\rm sinc}(x+y)$$ is continuous on all of ${\mathbb R}^2$. Since for $(x,y)\in D$ we have $f(x,y)=\tilde f(x,y)$ it follows that $$\lim_{(x,y)\to(2,-2)} f(x,y)=\lim_{(x,y)\to(2,-2)}\tilde f(x,y)=\tilde f(2,-2)=1\ .$$

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+1. I think I understand it better now. – anonymous Jan 22 '13 at 21:46