1) lim [x(y^2)] / (x^2 + y^2)
(x,y)->(0,0)
Find the value of the given limit, if it exists.

Using polar coordinates, set x = r cos(theta), y = r sin(theta)
Then, the given limit = lim [r cos(theta) r^2 sin^2(theta)] / r^2
r->0
= lim r [cos(theta) sin^(theta)]
r->0
= 0 since cos(theta) sin^(theta)<=1, i.e. bounded

If I found that the limit is equal to 0 this way, can I conclude immediately that the original limit is 0 too?

What I believe is that for the part using polar coordinates r->0, it seems that it's appraoching the origin through straight lines paths ONLY, however my textbook says that "for the limit to exist, we must get the same result no matter which of the infinite number of paths is chosen"


Thanks for answering!

The result is zero, regardless of the particular value theta, so the limit is zero. You're answer is fine.

The value of the function x^x as x\to0 is an example of a limit that doesn't exist because the value is path dependent. In fact, the value of x^x as can be made to approach any arbitrary value as x\to0 simply by choosing some appropriate path.

"The result is zero, regardless of the particular value theta, so the limit is zero. You're answer is fine." <---but this covers only the paths of all the straight lines through the origin, how about in the paths of parabolas and cube root function through the origin, etc?

As my textbook puts it, "for the limit to exist, we must get the same result no matter which of the infinite number of paths is chosen"

If you want to be extremely precise, use the epsilon-delta form. In multiple dimensions, \lim_{\vec x \to \vec x_0} f(\vec x) =L if there exists some \delta such that ||f(\vec x)-L|| < \delta\,\forall \vec x \in N_{\epsilon}(\vec x_0).

In this case, you have f(\vec x) = r\cos\theta\sin^2\theta in polar coordinates and L=0. In polar coordinates, the epsilon neighborhood of the origin is simply r < \epsilon. Thus you need to show that a delta>0 exists for each epsilon>0 such that |r\cos\theta\sin^2\theta| < \delta\; \forall r< \epsilon . Use the fact that \cos\theta\sin^2\theta is bounded.

kingwinner:

Could you please use TeX tags to post this kind of thing? It's much easier to read. For example:

\lim\limits_{(x,y)\rightarrow(0,0)} \frac{xy^2}{x^2 + y^2}

kingwinner:

Could you please use TeX tags to post this kind of thing? It's much easier to read. For example:

\lim\limits_{(x,y)\rightarrow(0,0)} \frac{xy^2}{x^2 + y^2}

Sorry, I still don't know how to use TeX.
But I will try to make my equations as clear as possible.

1) I am wondering whether using polar coordinates will cover ALL possible paths (straight lines, parabolas, etc.) to approach (0,0).

Say, if I am trying to eavluate a limit by changing it to polar coordinates, and I get a finite limit L using the polar coordinates, can I ALWAYS immediately conclude that the original limit is L in any case like this?

1) I am wondering whether using polar coordinates will cover ALL possible paths (straight lines, parabolas, etc.) to approach (0,0).

It seems like it should cover all continuous paths to (0,0), in that for sufficiently small distances from the origin, they'll all look more-or-less like straight lines. This wouldn't apply to discontinuous paths to the origin, but those are not really interesting in this context.

There is literally no way to follow all possible paths (this is a bigger infinity than the number of reals). However, you don't have to. Just use the squeeze theorem.