Question

Use polar coordinates to find the limit. If (r, θ) are polar coordinates of the point (x, y) with r ≥ 0, note that r → 0+ as (x, y) → (0, 0). (If an answer does not exist, enter DNE.) lim (x, y)→(0, 0) (x^6 + y^4) / (x^2 + y^2)

Answer 1

In polar coordinates, $x=r\cos\theta$ and $y=r\sin\theta$. So the limit is equivalent to

$\displaystyle\lim_{(r,\theta)\to(0,0)}\frac{(r\cos\theta)^6+(r\sin\theta)^4}{(r\cos\theta)^2+(r\sin\theta)^2}=\lim_{(r,\theta)\to(0,0)}r^2(r^2\cos^6\theta+\sin^4\theta)$

Since $-1\le\cos\theta\le1$ and $-1\le\sin\theta\le1$, we have $0\le\cos^6\theta\le1$ and $0\le\sin^4\theta\le1$, so that the behavior of $r^n$ as $r\to0$ decides the behavior of the overall function, and the limit would be 0.