Use polar coordinates to find the limit [Hint: Let and note that

Question

Use polar coordinates to find the limit  [Hint: Let  and note that

Quizplus · Accepted Answer

Let $r = \sqrt{x^2 + y^2}$ be the distance from the origin and $	heta$ be the angle that the point $(x,y)$ makes with the positive $x$-axis ($0 \leq 	heta < 2\pi$).
We have that $r^2 = x^2 + y^2 \geq xy$, so $r \geq \sqrt{xy}$. Taking the limit as $(x,y) 	o (0,0)$, we have
$$\lim_{(x,y)	o(0,0)} r\sin\left(\frac{\pi}{xy}ight) \leq \lim_{(x,y)	o(0,0)} r\sin\left(\frac{\pi}{xy}ight) \cdot \frac{\pi}{xy} \leq \lim_{(x,y)	o(0,0)} \frac{\pi r}{xy}.$$Since $\sin(\pi/x)$ is bound between $-1$ and $1$, we have $\lim_{(x,y)	o(0,0)} r\sin(\pi/xy) = 0$. Additionally, we have that $r 	o 0$ as $(x,y) 	o (0,0)$, so $\lim_{(x,y)	o(0,0)} \frac{\pi r}{xy} = \infty$. Therefore, by the squeeze theorem, $\lim_{(x,y)	o(0,0)} r\sin(\pi/xy) = \boxed{	extbf{(D)}\; 	ext{DNE}}$.

Use Polar Coordinates to Find the Limit [Hint: Let