# Give a $\delta-\epsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$.

I am new to multivariable limits and I am having a hard time arguing this limit:

Give a $$\delta$$-$$\epsilon$$ proof that $$f(x,y)=\sqrt{x+y-2}$$ is continuous at $$(2,1)$$.

I can work out the overarching context of the proof as follows: Let $$\epsilon>0$$ be given. We need a $$\delta$$ so that whenever $$\|(x,y)-(2,1)\|<\delta$$ we have $$\|f(x,y)-f(2,1)\|<\epsilon.$$ The difficulty I always have with these problems is that I know I need to rewrite $$\|\sqrt{x+y-2}-1\|$$ in terms of the given $$\delta$$ estimate but for some reason when it comes to these limit problems algebra turns into the most difficult thing in the world for me.

• Please show your working, in particular where you are stuck. You know that for all $\epsilon$ you need a $\delta$, so try working backwards. If you have not done any effort, then at least mention the definition of continuity, and write down what you have to prove in terms of that definition. – Teresa Lisbon Mar 1 '19 at 3:51
• You should be able to articulate more than just "I have this problem and no idea." In particular, what is the definition of continuity that applies to your bivariate function? How does it apply in this instance? The terse form of problem statement, without digesting what the problem means or the difficulty it entails, has resulted in the Question being placed on-hold for you to add context. – hardmath Mar 1 '19 at 16:33

## Break down the problem

This function is a compositition of the multivariate affine function $$(x, y) \mapsto x + y - 2$$, and the univariate square root function $$x \mapsto \sqrt{x}$$. Both of these functions are continuous on their domains. Affine functions are very straightforward functions, and are easier to prove continuous. Try first proving the continuity of this function. You may find it helpful to remember the triangle inequality and that \begin{align*} |x| &= \sqrt{x^2} \le \sqrt{x^2 + y^2} = \|(x, y)\| \\ |y| &= \sqrt{y^2} \le \sqrt{x^2 + y^2} = \|(x, y)\|. \end{align*}

The square root function is a little more complicated, but it is a single-variable function. Think about how to prove this function is continuous at $$x = 1$$ (or any $$x > 0$$). It involves some rationalisation, but hopefully if you're comfortable with $$\varepsilon$$-$$\delta$$ proofs in one variable, it shouldn't be an issue.

Then, putting it together, composing two continuous functions produces a continuous function. Think about the proof of this; essentially the $$\delta$$ in one function becomes the $$\varepsilon$$ in the other.

## Figure out the steps

As with most limit questions, your proof will be figured out in reverse order to the way it should be written. People often find it's easier to start with the conclusion $$|f(x, y) - f(x_0, y_0)| < \varepsilon$$, working backwards, until they reach $$\|(x, y) - (x_0, y_0)\| < \delta$$, for some positive number $$\delta$$. Remember, unlike in many other mathematics proofs, it's important that each step implies the previous step, not the next step. We want a full proof to start from the assumption $$\|(x, y) - (x_0, y_0)\| < \delta$$ and reach the conclusion $$|f(x, y) - f(x_0, y_0)| < \varepsilon$$, in other words, to be presented in the reverse order to how it was written.

So, let's begin with $$|\sqrt{x + y - 2} - 1| < \varepsilon$$. Note how this is reminiscent of the (univariate) square root function continuity proof. At this point, we may want to perform the same rationalisation trick. We have \begin{align*} |\sqrt{x + y - 2} - 1| &= |\sqrt{x + y - 2} - 1| \cdot \frac{|\sqrt{x + y - 2} + 1|}{|\sqrt{x + y - 2} + 1|} \\ &= \frac{\left|(\sqrt{x + y - 2})^2 - 1^2\right|}{\sqrt{x + y - 2} + 1} \\ &= \frac{\left|x + y - 3\right|}{\sqrt{x + y - 2} + 1}. \end{align*}

We are trying to make this expression small. By looking at the numerator, this is exactly $$|x + y - 2 - 1|$$, which is the expression you'd be making $$\varepsilon$$ small in the proof that $$(x, y) \mapsto x + y - 2$$ is continuous at $$(2, 1)$$. So, we shouldn't have any trouble making this numerator small.

The only thing we need to ensure is that our denominator doesn't become small in the process too. Fortunately, our denominator is $$1$$ plus a positive number; the denominator is at least $$1$$ (provided the expression is defined). So, for all such $$x, y$$, we have $$|\sqrt{x + y - 2} - 1| = \frac{\left|x + y - 3\right|}{\sqrt{x + y - 2} + 1} \le \left|x + y - 3\right|,$$ hence if we ensure $$\left|x + y - 3\right| < \varepsilon$$, then we have what we need.

There was the caveat that $$\sqrt{x + y - 2}$$ must be well-defined. This shouldn't be overlooked. If our $$\delta$$ is too large, then certain $$(x, y)$$ satisfying $$\|(x, y) - (2, 1)\| < \delta$$ may make $$f(x, y)$$ undefined. For example, if we had $$\delta = 3$$, then $$\|(0, 1) - (2, 1)\| = 2 < \delta$$, but $$f(0, 1)$$ is not defined. All we need to do is put a hard limit on how large $$\delta$$ can be, so that we are absolutely sure that none of these undefined points sneak in. We can see that $$\delta \le \frac{1}{2}$$ works, as $$\|(x, y) - (2, 1)\| < \frac{1}{2} \implies \begin{Bmatrix}|x - 2| < \frac{1}{2} \implies x > \frac{3}{2} \\ |y - 1| < \frac{1}{2} \implies y > \frac{1}{2} \end{Bmatrix} \implies x + y - 2 > \frac{3}{2} + \frac{1}{2} - 2 = 0.$$ (There are larger bounds on $$\delta$$ we could use, but $$\frac{1}{2}$$ works, so let's press on.)

Finally, we look at making $$|x + y - 3| < \varepsilon$$. If you could prove $$(x, y) \mapsto x + y - 2$$ is continuous, there should be no problem here. If you couldn't, apply triangle inequality on $$\Bbb{R}$$ (triangle inequality is a really, really useful tool in multivariate limits): $$|x + y - 3| = |(x - 2) + (y - 1)| \le |x - 2| + |y - 1|.$$ As above, we have $$|x - 2|, |y - 1| \le \|(x, y) - (2, 1)\|$$, so if we force $$\|(x, y) - (2, 1)\| < \frac{\varepsilon}{2}$$, we should be ok. So, we choose $$\delta = \frac{\varepsilon}{2}$$ provided $$\varepsilon$$ is small enough, but never let it be more than $$\frac{1}{2}$$. Our choice of $$\delta$$ is therefore $$\delta = \min\left\{\frac{1}{2}, \frac{\varepsilon}{2}\right\}.$$

## Compile the final proof

(We now write up all our steps in proper order. This bit is the only bit you need to write up when showing the proof to other people.)

Suppose $$\varepsilon > 0$$, and $$\|(x, y) - (2, 1)\| < \min\left\{\frac{1}{2}, \frac{\varepsilon}{2}\right\}.$$ Note that this implies that $$\|(x, y) - (2, 1)\| < \frac{1}{2} \implies \begin{Bmatrix}|x - 2| < \frac{1}{2} \implies x > \frac{3}{2} \\ |y - 1| < \frac{1}{2} \implies y > \frac{1}{2} \end{Bmatrix} \implies x + y - 2 > \frac{3}{2} + \frac{1}{2} - 2 = 0,$$ hence $$(x, y) \in \operatorname{Domain}(f)$$. Further, $$\|(x, y) - (2, 1)\| < \frac{\varepsilon}{2},$$ and so $$|x + y - 3| \le |x - 2| + |y - 1| \le 2\|(x, y) - (2, 1)\| < \varepsilon.$$ Therefore, $$|\sqrt{x + y - 2} - 1| = \frac{|x + y - 3|}{\sqrt{x + y - 2} + 1} \le |x + y - 3| < \varepsilon.$$ Hence, $$\lim_{(x, y) \to (2, 1)} \sqrt{x + y - 2} = 1$$.

Let $$\varepsilon > 0$$ and set $$\delta = \min\left\{\frac{\varepsilon^2}{6}, \frac12\right\}$$.

Let $$\|(x,y) - (2,1)\| < \delta$$. This implies $$|x-2| < \delta$$ and $$|y-1| < \delta$$. Also $$x+y-2 = 1 + (x-2) + (y-1) \ge 1-|x-2|-|y-1| > 1-2\delta \ge 0$$ so $$\sqrt{x+y-2}$$ is well-defined.

We have $$|x+y-2| \le |x-2| + |y-1| + 1 < 2\delta + 1$$ $$|x+y-2| \ge x+y-2> 1-2\delta$$

so \begin{align} \left(\sqrt{x+y-2}-1\right)^2 &= (x+y-2) - 2\sqrt{x+y-2}+1 \\ &< (2\delta+1) - 2\sqrt{1-2\delta} + 1 \\ &\le (2\delta+1) - 2(1-2\delta) + 1 \\ &\le 6\delta \\ &< \varepsilon^2 \end{align}

and hence $$\left|\sqrt{x+y-2}-1\right| < \varepsilon$$