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) x6 + y3 x2 + y2

Answer 1

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^6+y^3}{x^2+y^2}=\lim_{r\to0}\frac{(r\cos\theta)^6+(r\sin\theta)^3}{(r\cos\theta)^2+(r\sin\theta)^2}=\lim_{r\to0}\frac{r^6\pm r^3}{r^2}$

where in the last equality, the denominators of both limands are equal, but in the numerator we use the fact that $0\le\cos^6\theta\le1$ and $-1\le\sin^3\theta\le1$. In other words, the numerator is bounded above and below by polynomials in $r$. The numerator has degree higher than that of the denominator, so as $r\to0$, so is the limit equal to 0.