Use polar coordinates to find the limit of the function as (x, y) approaches (0, 0).
-$$f ( x , y ) = \frac { 9 x y } { \sqrt { x ^ { 2 } + y ^ { 2 } } }$$
A) $- 1$
B) 0
C) 1
D) $\pi$

Question

Quizplus · Accepted Answer

Using polar coordinates, we have $x=r\cos	heta$ and $y=r\sin	heta$. Substituting this into the function, we get
$$f(r,	heta)=\frac{9r^2\cos	heta\sin	heta}{r} = 9r\cos	heta\sin	heta.$$
We can see that $|f(r,	heta)|\leq 9r$, since $|\cos	heta|\leq 1$ and $|\sin	heta|\leq 1$. Therefore, by the squeeze theorem, we have
$$\lim_{(x,y)	o(0,0)} f(x,y) = \lim_{r	o 0} f(r,	heta) = \lim_{r	o 0} 9r\cos	heta\sin	heta = 0,$$
since $|\cos	heta\sin	heta|\leq \frac{1}{2}$ for any $	heta$.

Use Polar Coordinates to Find the Limit of the Function f(x,y)=9xyx2+y2f ( x , y ) = \frac { 9 x y } { \sqrt { x ^ { 2 } + y ^ { 2 } } }f(x,y)=x2+y2​9xy​

Use Polar Coordinates to Find the Limit of the Function $f ( x , y ) = \frac { 9 x y } { \sqrt { x ^ { 2 } + y ^ { 2 } } }$