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I was studying for some quizzes when I stumbled upon this question:

The function $$f(x,y)=\begin{cases} \frac{\sin (x-y)}{x-y} &x\neq y\\ h & x=y \\ \end{cases} $$

is continuous. What is the value of $h$?

My work:

I noticed that the answer found in my book is $h = 1,$ so I think I can get $h = 1$ by doing this:

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

because everybody knows that

$$\lim_{x\to0} \frac{\sin x}{x} = 1$$

Is my train of logic correct?

Sine of the Time
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fitzmerl duron
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    Not quite - you are interested not only in what happens when $x$ and $y$ approach $0$, but also what happens when they approach each other, i.e. what happens near the line $y = x$. – NickD Aug 10 '17 at 03:50

3 Answers3

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As Nick mentioned, the question is asking to find $h$ when $x=y$ rather than $\lim\limits_{(x,y)\to(0,0)}f(x,y)$. For example, this could be the case that $x=4$ and $y=4$.$$———————$$ This question becomes significantly simpler in polar coordinates: $x=r\cos(\theta)$, $y=r\sin(\theta)$.

$f(x,y)=\dfrac{\sin(x-y)}{(x-y)}\quad$ becomes

$f(r,\theta)=\dfrac{\sin[r(\cos\theta-\sin\theta)]}{r(\cos\theta-\sin\theta)}$.

Now, $\lim\limits_{(x,y)\to(0,0)}f(x,y)$ is equivalent to $\lim\limits_{\theta\to\frac{\pi}{4}}f(r,\theta)$.
Using L'hopital's rule, you get

$\dfrac{-r\sin\theta-r\cos\theta}{-r\sin\theta-r\cos\theta}=1\;$ at $\;\theta=\dfrac{\pi}{4} \space\space\forall{r}$.

Angelo
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rhdxor
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  • I like to be enlightened. What is the logic behind your answer? What is in $f(x,y) = \frac{sin (x-y)}{x-y}$ that we apply your solution that way? – fitzmerl duron Aug 10 '17 at 04:39
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    When dealing with vector spaces with dim > 3, I usually transform coordinates to simplify questions. In my experience, it is often the case that when global coordinates (x,y) make a problem difficult, curvilinear coordinates (r,theta) simplify the problem and vice versa for questions of these types. – rhdxor Aug 10 '17 at 04:56
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We could also say let $a=x-y$, then notice that as $x\to y$ then $a\to 0$. So find (as you have mentioned): $$\lim_{a\to 0}\frac{\sin(a)}{a}.$$

Alternatively, we can write $$\lim_{x\to y}\frac{\sin(x-y)}{x-y}=\frac{0}{0}$$ In this case, L'Hospital's rule can be applied to find $$\lim_{x\to y}\frac{\sin(x-y)}{x-y}=\lim_{x\to y}\frac{\frac{\partial}{\partial x}\sin(x-y)}{\frac{\partial}{\partial x}(x-y)}=\lim_{x\to y}\frac{\cos(x-y)}{1}=\cos(0)$$

MattW
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Correct me if wrong :

$f(x,y) = f(x-y) = \frac{\sin(x-y)}{x-y)}$, $x \ne y$.

Set $z: = x-y$.

$f(z) = \frac{\sin(z)}{z}$, $ z \ne 0$.

$f$ is continuous at $z = 0$:

$\star)$ $\lim_{z \rightarrow 0} f(z) = f(0)$.

$\lim_{z \rightarrow 0} \frac{\sin(z)}{z} = 1$.

$\star$): continuity at $z=0$ implies:

$f(0) := 1$.

Peter Szilas
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