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))((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.