Consider
$$f(x,y) =\begin{cases} \frac{xy}{\sqrt{x^2 +y^2}
} & (x,y)\neq (0,0) \\
0 & (x,y)=(0,0)
\end{cases}$$
Show that $f(x,y)$ is continuous at $(0,0)$.
My method
We have,
$|f(x,y)-0| = |x||\frac{y}{\sqrt{x^2 +y^2}}| < \delta\cdot1$.
Consider, $\delta<\varepsilon$. ($\delta>0$)
Then for any $\varepsilon > 0$, there exists a $\delta<\varepsilon$ such that $|f(x,y)-0|<\varepsilon$ for all $(x,y)$ that satisfies the inequation $x^2 +y^2 < \delta^2$. Hence, $f(x,y)$ is continuous at $(0,0)$.
Is this proof correct?
Best Answer
Your choice of $\delta$ works. Indeed, choosing $\delta = \varepsilon$ works as well; don't forget, if you were to choose this, then $\sqrt{x^2 + y^2} < \delta$ if and only if $\sqrt{x^2 + y^2} < \varepsilon$.
What's not so clear from your proof is why this choice works. You need to work on your mathematical exposition. The last we see any specifics about $f(x, y)$ is the first line. After this, the choice of $\delta$ is merely asserted. What is it about this function $f(x, y)$ that makes this choice of $\delta$ appropriate?
Here's an argument that I think your proof is missing. Note that $$x^2 + y^2 \ge y^2$$ as $x^2 \ge 0$. Taking the square root of both sides, $$\sqrt{x^2 + y^2} \ge |y|,$$ hence $$\left|\frac{y}{\sqrt{x^2 + y^2}}\right| \le 1 \implies |x| \left|\frac{y}{\sqrt{x^2 + y^2}}\right| \le |x| \le \sqrt{x^2 + y^2},$$ by a similar argument as above, involving $x$ instead of $y$. Thus, if we assume $0 < \sqrt{x^2 + y^2} < \varepsilon$, then $$|x| \left|\frac{y}{\sqrt{x^2 + y^2}}\right| < \varepsilon,$$ which implies $|f(x, y) - 0| < \varepsilon$.