Use polar coordinates to find the limit of the function as (x, y) approaches (0, 0).
-$$f ( x , y ) = \cos ^ { - 1 } \left( \frac { x ^ { 3 } - x y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } } \right)$$
A) $\pi$
B) $\frac { \pi } { 2 }$
C) 1
D) No limit

Question

Quizplus · Accepted Answer

In polar coordinates, we have $x = r\cos	heta$ and $y = r\sin	heta$. Substituting these into the given function and simplifying, we get 
$$f(r,	heta) = \cos^{-1}\left(\frac{r^2\cos^3	heta - r^3\sin^2	heta\cos	heta}{r^2}ight) = \cos^{-1}(\cos^3	heta - r\sin^2	heta\cos	heta)$$
Since $\cos^{-1}$ is a continuous function, we can find the limit as $(r,	heta)$ approaches $(0,0)$ by plugging in $r=0$ to the above expression. This gives us
$$\lim_{(r,	heta)ightarrow(0,0)} f(r,	heta) = \cos^{-1}(\cos^3 0 - 0\sin^2 0\cos 0) = \cos^{-1}(1) = \boxed{\frac{\pi}{2}}$$

Use Polar Coordinates to Find the Limit of the Function f(x,y)=cos⁡−1(x3−xy2x2+y2)f ( x , y ) = \cos ^ { - 1 } \left( \frac { x ^ { 3 } - x y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } } \right)f(x,y)=cos−1(x2+y2x3−xy2​)

Use Polar Coordinates to Find the Limit of the Function $f ( x , y ) = \cos ^ { - 1 } \left( \frac { x ^ { 3 } - x y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } } \right)$