Is this a valid proof of the continuity of $f(x)=\sqrt{x}$ at $x=2$, using epsilon-delta definition?

Question

Prove that $f(x)=\sqrt{x}$ is continuous at $x=2$

Proof:

Givens: $\varepsilon > 0$ and $\delta > 0$ and $|x-2| < \delta$ for $\delta < 1$. Let $0 < \delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$ $$ |f(x)-f(2)| = |\sqrt{x}-\sqrt{2}| < \varepsilon \tag1 $$ $$ x-2 < |\sqrt{x}+\sqrt{2}|\varepsilon \tag2$$ Also, using the givens, $$ |x-2| < \delta < 1 \implies |x-2| < 1 \implies 1 < x < 3 \tag3 $$ Then it follows that $$ x - 2 < |\sqrt{3} + \sqrt{2}|\varepsilon \tag4 $$ Therefore, $$ |f(x)-f(2)| < \frac{\delta}{\sqrt{3}+\sqrt{2}} < \varepsilon. \tag5 $$ The reason I am unsure is because I know that delta should be minimized but I am unsure if all my steps are correct.

Answer 1

A general way to choose $\delta$ for any invertible function is the following: You want to show that for each $\epsilon>0$ there is $\delta>0$ such that, for each $x \in (2-\delta, 2+\delta)$ $$ \sqrt{2}-\epsilon<\sqrt{x}<\sqrt{2}+\epsilon $$ If $\epsilon>\sqrt{2}$ you can verify that the inequality is satisfied for any $\delta < (\epsilon+\sqrt{2})^2-2$. So we can assume $\epsilon \le \sqrt{2}$. By squaring all the members (that are all non-negative) $$ (\sqrt{2}-\epsilon)^2<x<(\sqrt{2}+\epsilon)^2 $$ So you want $$ 2-\delta >(\sqrt{2}-\epsilon)^2 $$ $$ 2+\delta < (\sqrt{2}+\epsilon)^2 $$ i.e. $$ \delta < \epsilon(2\sqrt{2} -\epsilon) $$

Answer 2

For me you proof is very confusing and it is wrong. It is wrong because the equation $|f(x)-f(2)| < \frac{\delta}{\sqrt{3}+\sqrt{2}}$ does not hold. $|f(2.1)-f(2)|\approx 0.034$, but $\frac{0.1}{\sqrt3+\sqrt2}\approx 0.032$.

But now to the confusion. You state

Givens: $\varepsilon > 0$ and $\delta > 0$ and $|x-2| < \delta$ for $\delta < 1$.

and then you state

Let $0 < \delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$

But what is the difference? The inequalities $0 < \delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$ is also given. And if $\delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$ then $\delta < 1$ is redundant. So you can leave out this statement. Finally you do not have to prove that there is an interval of $\delta$ where your statment holds, it is sufficient that for each $\varepsilon$ you find on $\delta$. From you inequalities I deduce that $\min \{\frac12, \varepsilon\}$ is such a $\delta$, so I would write

Given $\varepsilon>0$ choose $\delta=\min{\frac12, \varepsilon}$ and assume that $|x-2|<\delta$.

Then you state:

$$|f(x)-f(2)| = |\sqrt{x}-\sqrt{2}| < \varepsilon \tag1$$ $$ x-2 < |\sqrt{x}+\sqrt{2}|\varepsilon \tag2$$

And what's that? I think you want to tell us

We have to show: $$|f(x)-f(2)| = |\sqrt{x}-\sqrt{2}| < \varepsilon \tag1$$

I think you want to tell us that and not that this is a prerequisite. So I think you should state this explicitly.

An what is $(2)$? We have

$$(\sqrt x-\sqrt 2)(\sqrt x+\sqrt 2)=(x-2)$$

This is formula is crucial to the proof so I would explicitly mention it explicitly.

From this and $(1)$ follows $$ x-2 < |\sqrt{x}+\sqrt{2}|\varepsilon \tag2$$

$(3)$ and $(4)$ follow immediately from the definition of $\delta$ and the $|x-2|<\delta$. I do not understand why you write "Then it follows" to justify $(4)$. Because it does not follow from $3$ Finally you do not tell us how you get inequality $(5)$. As I mentioned at the beginning. This inequality is wrong.

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Here is a short proof:

Assume that $\varepsilon>0$ and $\delta=\varepsilon.$ If $|x-2|<\delta$ then $$|f(x)-f(2)|=|\sqrt x-\sqrt 2|=\frac{|\sqrt x-\sqrt 2|\cdot|\sqrt x+\sqrt 2|}{|\sqrt x+\sqrt 2|}=\frac{|x-2|}{\sqrt x+\sqrt 2}\lt\frac{\delta}{\sqrt 2}<\varepsilon$$ So $f(x)=\sqrt x$ is continuous at $x=2.$

Answer 3

You can prove something a bit stronger with less work.

$f(x)=\sqrt{x}$

$\sqrt{x}-\sqrt{c}= \frac{x-c}{\sqrt{x}+\sqrt{c}}$

WLOG, $x>c$

$x,c>1\implies \sqrt{x}-\sqrt{c}<(x-c)/2$

$|x-c|<\epsilon\implies \sqrt{x}-\sqrt{c}<\epsilon/2$

So $\delta=\epsilon$ proves uniform continuity in the region in question $(1,\infty)$.

Uniform continuity implies continuity and $c=2$ is in that region where the function is uniformly continuous, so the function is continuous at $x=2$.