Use polar coordinates and L'Hopital's Rule to find the limit.

Question

Quizplus · Accepted Answer

Using polar coordinates, the limit becomes:

$$\lim\limits_{(x,y)	o(0,0)}\frac{x^2y}{x^4+y^2}=\lim\limits_{r	o0}\frac{r^3\sin	heta\cos^2	heta}{r^4\cos^4	heta+r^2\sin^2	heta}=\lim\limits_{r	o0}\frac{r\sin	heta\cos^2	heta}{r^2\cos^4	heta+\sin^2	heta}$$

Note that $\frac{r\sin	heta\cos^2	heta}{r^2\cos^4	heta+\sin^2	heta}$ is of the form $\frac{0}{0}$ as $rightarrow 0$. Therefore, we may apply L'Hopital's Rule. Taking the derivative of the numerator and denominator with respect to $r$, we get:

$$\lim_{r	o0}\frac{\sin 2	heta}{4r^3\cos^6	heta}=\frac{\sin 2	heta}{0}$$

As $rightarrow 0$, the denominator goes to $0$ much faster than the numerator, so the limit is $0$.

Thus, the answer is B: 11ebbd4d_3f09_a4a3_8a0a_6b043c28fede_TB8202_11.

Use Polar Coordinates and L'Hopital's Rule to Find the Limit