# Can we Conclude Limit is zero from Polar Co-ordinates

I am trying to Evaluate the Limit: $$L=\lim_{(x,y)\to (0,0)}\frac{x^3+y^3}{x^2+2y}$$

I tried the Paths: $$y=x^2$$, $$y=mx$$, etc i got the Limit as Zero. Finally I used Polar Coordinates we get: $$L=\lim_{r \to 0}\frac{r^2(\cos^3 t+\sin^3 t)}{r\cos^2 t+2\sin t}=0$$

I have doubt that does Polar Co-ordinates covers all the Paths towards Origin or only Straight line Paths?

EDIT: Now based on this i have gone through some existing threads. I came to know that Polar coordinates are used to take path towards origin on a straight line only as $$\theta$$ is a constant thereof.

If we take any curvy path $$\theta=\theta(r)$$ and hence we cannot estimate the limit.

I tried Parabola path $$y=x^2$$ in Polar coordinates which is:

$$r=\sec(\theta)\tan(\theta)$$

Expressing $$r$$ in terms of $$\theta$$ we get:

$$\tan(\theta)=\frac{-1+\sqrt{1+4r^2}}{2}$$

Now as $$r \to 0$$ $$\theta \to 0$$ is this possible?

• If you can use squeeze theorem, polar coordinates will cover everything, but I don't see an immediate manipulation to make that happen. – Ninad Munshi Jan 3 '20 at 17:57

If both numerator and denominator are homogeneous, polar coordinates can finish the problem. Here, the denominator is $$x^2 + 2 y,$$ and this is exactly equal to zero along the parabola $$y = -\frac{x^2}{2}$$ which passes through the origin.
OR: along the path $$y = - \frac{x^2}{2} + x^3,$$ your function (the fraction) has limit $$\frac{1}{2}\; . \; \;$$ along the path $$y = - \frac{x^2}{2} - x^3,$$ your function (the fraction) has limit $$\frac{-1}{2}$$
For an explicit example of a limit that approaches a nonzero, noninfinite value, take $$y = -\frac{1}{2}x^2 + x^3$$
$$\lim_{x\to 0} \frac{x^3-x^6(\frac{1}{2} - x)^3}{x^2 - 2(\frac{1}{2}x^2 - x^3)} = \lim_{x\to 0} \frac{x^3 + O(x^6)}{2x^3} = \frac{1}{2}$$