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I'm learning to take limits of functions $f : \mathbb{R}^2 \to \mathbb{R}$. I know the formal definition, but AFAICT, that is too rigorous and complex to use in actual examples.

I also know the following methods: finding paths in which the limits are different (to prove it doesn't exist), and, if I can't find, I need to use the Squeeze Theorem to find functions that limit the value of mine. I also know a technique (that works when $(x, y) \to(0, 0)$) that I'm pretty sure is derived from the previous, in which I must write $f$ as a product of a limited function and a function that goes to zero (if I can, the limit is zero). The problem is, if I can't find that, and I'm pretty sure the limit exists, what should I do to find it?

Is the formal definition the only way?

For example, take this limit:

$\lim_{(x, y) \to(0, 0)} \frac{xy}{y - x^3}$

I've been struggling with it for a few days, and none of the methods I mentioned seem to work...

Luan Nico
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2 Answers2

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One possible approach:

First, we can see the degree and naively guess numerator would go to zero faster than denominator so that limit would be zero.

Next, we can divide some variable for clarification, like

$\left|\frac{xy}{y-x^3}\right| = \left|\frac{x}{1-x^3/y}\right|$.

Lastly we can check by the different path method you mentioned.

so that if $y = x^3$, This function cannot be defined.

If $y = k x^3 (k \not=1)$, $\left|\frac{x}{1-x^3/y}\right| = \left|\frac{x}{1-1/k}\right|\to0$.

If $x<1$ and $1-x^3/y = x$, which means $y = x^3/(1-x)$, $\left|\frac{x}{1-x^3/y}\right| = \left|\frac{x}{x}\right| = 1$, so that the limit does not exist.

Arch
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  • Nice catch with the division and $y = \frac{x^3}{1 - x}$, I wouldn't have though of that in ten years! :) – Luan Nico Apr 03 '15 at 14:05
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Do you know this theorem, which holds in $\mathbb{R}^2$:

$(x_n,y_n) \rightarrow (x,y)$ iff $x_n \rightarrow x$ and $y_n \rightarrow y$

ben300694
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