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

Solution 1:

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

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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.

Solution 2:

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:

Solution 3:

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 $$\lim_{(x,y)\to(2,-2)} f(x,y)=\lim_{(x,y)\to(2,-2)}\tilde f(x,y)=\tilde f(2,-2)=1\ .$$