Proof: $f(x,y)=\sqrt{4x^2+y^2}$ is continuous at $(0,0)$

Question

Prove $f(x,y)=\sqrt{4x^2+y^2}$ is continuous at $(0,0)$.

Attempt

I need to find a $\delta(\epsilon)$: $$\forall \epsilon>0\exists \delta>0: 0<\sqrt{x^2+y^2}<δ \implies |\sqrt{4x^2+y^2}|<\epsilon $$

I set $\delta=\dfrac \epsilon 2$ so $$x^2+y^2<\dfrac {\epsilon^2}4 \implies 4x^2+y^2\leq 4x^2+4y^2< \epsilon^2 \implies f(x,y)=\sqrt{4x^2+y^2}<\epsilon$$

The only thing missing is the absolute value, but as I'm taking the square root of squares (I assume real numbers, problem doesn't mention it), $f \geq 0$ so $f=|f|$ so I'm done, is this correct?

A more important question, if this is correct, is how would you find a value for $\delta$? I did this by trying out many values, but don't know how to arrive at an expresion $\delta =g(\epsilon)$ in a "deductive" fashion.

E: Forgot to add this when I posted this question:

Is it possible to generalize this method to prove continuity of this function in $\Bbb R^2$?

Answer 1

The proof is fine as-is. To answer your latter question of how to find $\delta(\varepsilon)$, you typically work backwards. Force the difference between the function and the limit to be less than $\varepsilon$, and then use algebraic manipulations (or casework) to get the difference to "look like" the distance function.

In your problem, we must force $$\left|\sqrt{4x^2 + y^2} \right| < \varepsilon.$$ Similar to how you applied your value of $\delta$ in the proof, we notice that this almost looks like $\sqrt{x^2 + y^2}$. To fix this, we can bound the absolute value by a larger value that looks more like the distance function. That is, $$\left|\sqrt{4x^2 + y^2} \right| < \left|\sqrt{4x^2 + 4y^2} \right| < \varepsilon.$$ Simplifying the middle expression gives us $2 \sqrt{x^2 + y^2}$, which we assume to be less than $2\delta$ (remember, working backwards). Dividing by 2 gives $\delta = \varepsilon/2$. Using $\delta-\varepsilon$ in multivariable limits is rather difficult in most cases. You will frequently see the squeeze theorem (which is effectively as rigorous as $\delta-\varepsilon$) or conversion to polar coordinates.

Answer 2

Your solution is correct. And as you pointed out it implicitly assumes that $x,y$ are real. This is because your definition of the limit already did, when you used "$0 < \sqrt{x^2+y^2} < δ$". If you want a more general definition of limit that works for complex $x,y$, then it goes as follows: $\def\cc{\mathbb{C}}$

$f \colon \cc^2 \to \cc$ is continuous at $(a,b) \in \cc^2$ iff:

  For any $ε > 0$:

    For some $δ > 0$:

      For any $z,w \in \cc$:

        If $|z-a| < δ$ and $|w-b| < δ$:

          $| f(z,w) - f(a,b) | < ε$.

[Note that if you apply my definition to the real case, the window is square instead of circular like in your definition, but square windows are easier to work with!]

To see what $δ$ works, you should not work from the end result you want. Rather just see what happens for arbitrary $δ$ and every time you want to restrict it to a smaller window around the point to get more information you can feel free to do so.

In the case of circular windows as in your definition, note that $\sqrt{x^2+y^2} < δ$ implies $x < δ$ and $y < δ$ (which is a square window, yay!), so we can just use that instead. So we have $\sqrt{4x^2+y^2} \le \sqrt{4δ^2+δ^2} = \sqrt{5} δ$, and we know what $δ$ to pick to make sure that the last expression is less than $ε$.

Answer 3

I would decompose

$x^2 = x \times x$ so it is continuous $y^2$ is also continuous.

so $$ 4x^2 + y^2 $$ is continuous as a linear sum of continuous functions.

We now show $$ f(z) = |z|^{1/2} $$ is continuous. I leave this to you but it is straightforward.

Now use the fact that the composition of continuous functions is continuous.

This is how an analyst would actually proceed rather working out an epsilon-delta argument.