multivariable limit problem involving sine This is a homework problem, but I am having lots of trouble figuring this out. I know homework problems aren't well accepted on this website, but I couldn't find a better place to ask this.
$$\lim_{(x,y)\to (0,0)}\sin\left(\frac xy\right)(ax+by)$$
Is there a way to bring the $ax+by$ inside the sine? I'm not so good with trig. But, really the $a,b$ are throwing me off. I'm not even sure if they are integers or what. The book doesn't say anything.
 A: Note that
$$-|ax+by|\leq \sin\left(\frac{x}{y}\right)(ax+by)\leq|ax+by|$$
Therefore, by the Squeeze Theorem, $$\lim_{(x,y)\to (0,0)}\sin\left(\frac {x}{y}\right)(ax+by)=0$$
A: Your function is not defined on the $x$-axis. Outside of that, no matter what the constants $a,b$ are, you have a bounded function times a function that approaches $0.$ What else can it do except go to $0?$
A: $a$ and $b$ are arbitrary real numbers in your problem.
One way to approach this is to see what happens if you approach $(0,0)$ along some arbitrary line.  If the answer for the limit is not independent of the direction of that line, then the limit does not exist.  So write your line as $y=mx$ (no constant; it's got to go through the origin).  (We will have to deal with the case of the line $x=0$ separately to make our set of lines complete.)
Then you have $$f(x,y) = (ax+bmx)\sin \left( \frac{x}{mx} \right)
= (ax+bmx) \sin \left( \frac{1}{mm} \right)$$ which clearly approaches zero as $x \to 0$.
But now look at the line $x=0$, that is, approaching the origin along the $y$ axis. 
$$f(0,y) = (by)\sin \left( \frac{0}{y} \right) $$ which of course is zero everywhere along that line.  
So it looks like your limit is zero.
