We need some $\epsilon-\delta$ here

$-|px+qy|\leq (px+qy)\sin\left(\frac{x}{y}\right)\leq |px+qy|$ so the limit is $\boxed{0}$ by the squeeze theorem .

Let $(r,\theta)$ be the polar coordinates of the point $(x,y)$ . Then $x = r\cos(\theta)$ and $y = r\sin(\theta),$ and since $r = \sqrt{x^{2} + y^{2}}$ it is clear that $r \rightarrow 0$ as $(x,y) \rightarrow (0,0).$ The limit then becomes

$\lim_{r \rightarrow 0} r(p\cos(\theta) + q\sin(\theta))\sin(\cot(\theta)) = rA(\theta),$

where $A(\theta) = (p\cos(\theta) + q\sin(\theta))\sin(\cot(\theta)).$

Now since both the sine and cosine functions have a range of $[-1,1]$ , we have that

$-(p + q) \le A \le (p + q) \Longrightarrow -r(p + q) \le rA \le r(p + q),$ as $r \ge 0.$

Then as $p$ and $q$ are both constants, by the Squeeze theorem we have that $\lim_{r \rightarrow 0} rA = \boxed{0}.$

Let the infinitesimal small value x attains be $h_{1}$ & that y attains be $h_{2}$ . Clearly, $sin \ (h_{1}$ / $h_{2})$ is a finite value lying between -1 & 1 .

Once that's established, we can clearly conclude 0*( a finite value) = 0. Note that the first part of the limit evaluates to an infinitesimal small value tending to zero.