Can we Conclude Limit is zero from Polar Co-ordinates

Question

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?

Answer 1

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.

Therefore the fraction cannot describe a continuous function. It is not defined along the parabola, and it has arbitrarily large values when we look at points very close to the parabola.

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 $\frac{-1}{2}$

Answer 2

For an explicit example of a limit that approaches a nonzero, noninfinite value, take $y = -\frac{1}{2}x^2 + x^3 $

$$\lim_{x\to 0} \frac{x^3-x^6(\frac{1}{2} - x)^3}{x^2 - 2(\frac{1}{2}x^2 - x^3)} = \lim_{x\to 0} \frac{x^3 + O(x^6)}{2x^3} = \frac{1}{2}$$