Limit of $\frac{x^2}{x^2 + y^2}$ as $(x,y) \to (0,0)$

Question

I am very new to multivariate calculus and I came up for an incorrect "proof" that $$\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^2} = 0.$$ I understand that the limit does not exist because approaching with $x=0$ and with $y=0$ yield different answers, however after several proofreads I do not understand where I have gone wrong with the "proof" given below.

We wish to show that for every $\epsilon$ there exists a $\delta$ such that $$\sqrt{x^2+y^2} < \delta \implies | \frac{x^2}{x^2 + y^2} | < \epsilon.$$ Choose $\delta = \min({\sqrt{\epsilon}, 1}).$ Now, because $\delta$ is at most 1,

$$\sqrt{x^2+y^2} < \delta \leq 1$$ $$1 < \frac{1}{x^2+y^2}$$ $$\sqrt{x^2+y^2} < \delta \implies \frac{x^2}{1} < \epsilon.$$

Substituting in $\delta = \sqrt{\epsilon}$, we obtain

$$x^2+y^2 < \epsilon \implies x^2 < \epsilon$$ which is true.

Answer 1

First off, I'd like to object to the choice of words "Substituting in $\delta = \sqrt{\varepsilon}$". You can't make this substitution as $\delta$ may be equal to $1$, not $\sqrt{\varepsilon}$. However, you know that $\delta \le \sqrt{\varepsilon}$, so $$\sqrt{x^2 + y^2} < \delta \le \sqrt{\varepsilon} \implies x^2 + y^2 < \varepsilon,$$ as required.

Aside from that, everything you've written is fine. However, you haven't completed the (necessarily false) proof! You've concluded two statements: \begin{align*} \frac{1}{x^2 + y^2} &> 1 \\ x^2 &< \varepsilon. \end{align*} These two correct statements do not imply that $\frac{x^2}{x^2 + y^2} < \varepsilon$. You can't simply multiply the inequalities, as they point in opposite directions. Play around with some specific values, and you'll see how the first two inequalities can be satisfied, but not the third.