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Can we make $\frac{\sin (x+y)}{x+y}$ continuous, defining it appropriately at $(0, 0)$ ??

What does it mean to make a function continuous??

N. F. Taussig
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Mary Star
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    You only want to define it at $(0,0)$? Or do you want to define it at all $(x,-x)$? – Thomas Andrews Mar 26 '15 at 23:25
  • Simply put, a continuous function $f(x)$ means that there are no breaks or cuts in the curve $y=f(x)$ – Prasun Biswas Mar 26 '15 at 23:26
  • The exercise asks at $(0, 0)$. But what does it mean to make a function continuous?? @ThomasAndrews – Mary Star Mar 26 '15 at 23:26
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    The question is whether $\lim_{(x,y) \to (0,0)} \frac{\sin(x+y)}{x+y}$ exists. If it does, then you can define a function which is equal to the given one where that is defined and equal to that limit at $(0,0)$. (Of course there is still concern about, e.g., $(1,-1)$.) – Ian Mar 26 '15 at 23:27
  • But what does it mean to MAKE a function continuous?? @PrasunBiswas – Mary Star Mar 26 '15 at 23:27
  • An informal version is, "A function $f(x)$ is continuous if you can draw the curve $y=f(x)$ in a single go without lifting your pen from the plane of paper. – Prasun Biswas Mar 26 '15 at 23:28
  • Well, making a function continuous can be thought of as finding suitable values of unknown constants in a function such that the limit definition of continuity holds. – Prasun Biswas Mar 26 '15 at 23:30
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    You make a function (take for instance $g(x) = \frac{x^3 + y^3}{x^2 + y^2}$) continuous by writing down a piecewise definition of it, like this: $$ g(x) = \cases{\frac{x^3 + y^3}{x^2 + y^2} & for $(x, y) \neq (0, 0)$\0 & for $(x, y) = 0$} $$ but in your case you have a whole line of singularities to worry about, not just the origin. – Arthur Mar 26 '15 at 23:30

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$\newcommand{\Reals}{\mathbf{R}}$Calculus books often play fast and loose with domains (and targets), but it's important to realize (and remember) that changing the domain of a function, even by adding or removing a single point, gives a different function. This is a mildly pedantic answer, but taking extra care may be worthwhile. Here are some relevant definitions:

If $X$ is a set, $A \subset X$, and $f:A \to \Reals$ is a function, then an extension of $f$ to $X$ is a function $F:X \to \Reals$ such that $$ F(x) = f(x)\quad\text{for all $x$ in $A$.} $$ (In other words, the restriction of $F$ to $A$ is $f$.)

A continuous extension of $f$ to $X$ is exactly what it sounds like: An extension of $f$ to $X$ that is, in addition, continuous.

Here $A = \{(x, y)\text{ in } \Reals^{2} : x + y \neq 0\}$ is the plane $\Reals^{2}$ with the line $y = -x$ removed; $X = A \cup \{(0, 0)\}$ is the set $A$ with the origin appended, and $f:A \to \Reals$ is defined by $$ f(x, y) = \frac{\sin(x + y)}{x + y}. $$ The question amounts to: Does there exist a continuous extension of $f$ to $X$? Informally, "can $f(0, 0)$ be defined in such a way that $f$ is continuous at $(0, 0)$?"

The answer in this example boils down to, "Does $\lim\limits_{(x, y) \to (0, 0)} f(x, y)$ exist?" If "yes", and if the limit is $\ell$, then defining $f(0, 0) = \ell$ gives a continuous extension of $f$ to $X$. If "no", $f$ has no continuous extension to $X$.

(Incidentally, as a couple of commenters have noted, the origin is not the only "problem point" for $f$. Is it possible that $$ f(x, y) = \frac{\sin(x^{2} + y^{2})}{x^{2} + y^{2}}\quad\text{or}\quad f(x, y) = \frac{\sin\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2}}} $$ instead? The latter two functions are defined everywhere except the origin.)

Andrew D. Hwang
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A function is continuous (in a domain) if it is continuous at every point of its domain. Saying this is equivalent as to saying that $$\lim_{\textbf {x} \to a} f(\textbf x) = f(\textbf a)$$ for every $a \in Dom f$.

Once you have this in mind, you need to estabilish your domain, and see that your function fulfills the requirements.

Aaron Maroja
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A function is continuous if for every $x$ in it's domain, it is true that: $$ \lim_{y\to x}f(y)=f(x) $$ This is the same as the curve being drawable with a pencil.

To make a function continuous, would probably mean that the function is not defined at, say $0$ because of division by zero, and then you figure out which value you could define the function to have at that point, such that the resulting function is continuous.

Alice Ryhl
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  • Pedantically speaking, it's not really possible to draw a non-rectifiable curve with a pencil... – Ian Mar 26 '15 at 23:51
  • Regarding your second claim, the "drawable with a pencil" criterion is actually technically wrong. In addition, a function of two variables (as in the OP's question) can "have its graph in one piece" and still be discontinuous. (That isn't the case here, but it's a logical possibility at the outset.) :) – Andrew D. Hwang Mar 26 '15 at 23:52