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I am currently undergoing a Calculus I course as an undergraduate. I was hoping to gain some guidance into whether the proof I have worked out for the continuity of a function $ f(x) = \sqrt{x} $ is robust enough or is missing out on certain details.

Aim: Prove that the function $f(x) = \sqrt{x}$ is continuous at $a > 0$.

(1) Finding the value of f(a) for which f is defined

The function $f(x) = \sqrt{x}$ is defined for values of $x \ge 0$. Given $a > 0$,

$$ f(a) = \sqrt{a} \\ \lim_{x \to a} \sqrt{x} = \sqrt{a} $$

(2) Showing that $\lim\limits_{x \to a} \sqrt{x}$ exists for every $a > 0$.

Hence, given $\epsilon > 0$, we aim to find a proper $\delta > 0$ such that:

$$0 < |x - a| < \delta \Rightarrow |\sqrt{x} - \sqrt{a}| < \epsilon$$

Factorizing $|x-a|$,

$$ |x-a| = |\sqrt{x} - \sqrt{a}||\sqrt{x} + \sqrt{a}| < \delta \\ |\sqrt{x} - \sqrt{a}| < \frac{\delta}{|\sqrt{x} + \sqrt{a}|} $$

Since $\sqrt{x} + \sqrt{a}\ge 0$, it follows that $\frac{1}{|\sqrt{x} + \sqrt{a}|} < 1$. Hence,

$$ |\sqrt{x} - \sqrt{a}| < \frac{\delta}{|\sqrt{x} + \sqrt{a}|} < \delta $$

Therefore, we take $\delta = \epsilon$.

Proof:

$$ 0 < |x - a| < \delta \Rightarrow |\sqrt{x} - \sqrt{a}| < \delta = \epsilon. $$

Since $f(a) = \sqrt{a}$ and $\lim\limits_{x \to a} \sqrt{x} = \sqrt{a}$ for $a > 0$,

$$ \lim_{x \to a} \sqrt{x} = f(a) $$

and thus the function $f(x) = \sqrt{x}$ is continuous at every a > 0.

Any feedback on gaps or loopholes in this proof would be greatly appreciated!

Edit:

There was a flaw in my logic, as kindly pointed out below. Since $\sqrt{x} + \sqrt{a}$ may fall in the range of values $0 \le \sqrt{x} + \sqrt{a} \le 1$ the assertion that $\frac{1}{|\sqrt{x} + \sqrt{a}|}$ does not hold.

The appropriate logic should be:

$$ \frac{\delta}{|\sqrt{x} + \sqrt{a}|} < \frac{\delta}{|\sqrt{a}|} $$

for x > 0.

Hence,

$$ |\sqrt{x} - \sqrt{a}| < \frac{\delta}{|\sqrt{a}|} $$

We choose $\delta = \epsilon\sqrt{a}$.

Revised proof:

$$ 0 < |x - a| < \delta \Rightarrow |\sqrt{x} - \sqrt{a}| < \epsilon\sqrt{a}(\sqrt{a}) = \epsilon $$

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  • $\begingroup$ Close. But $\sqrt{x} + \sqrt{a} \geq 0$ does not imply $1/|\sqrt{x} + \sqrt{a}| < 1$. $\endgroup$ – aschepler Aug 22 '20 at 12:13
  • $\begingroup$ RIGHT! Thank you! Must have messed up it up by assuming that $a \ge 1$. $\endgroup$ – iobtl Aug 22 '20 at 13:14
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There is no reason why $\frac {\delta} {\sqrt x+\sqrt a}$ should be less than $\delta$ so your proof is not valid.

Note that $\frac {\delta} {\sqrt x+\sqrt a} < \frac {\delta} {\sqrt a}$ (for $x>0$) so it is enough to choose $\delta =\epsilon \sqrt a $

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  • $\begingroup$ I realize my mistake. Thank you for the correction! $\endgroup$ – iobtl Aug 22 '20 at 13:15
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Let $\varepsilon>0$ be defined. We want to find such $\delta>0$ that suffices $\forall x\in (a-\delta,a+\delta), |f(x)-f(a)|<\varepsilon$. So, we're looking for such $\delta$ that: $|\sqrt x-\sqrt a|<\varepsilon$. Notice: $|\sqrt x-\sqrt a|=\frac{|x-a|}{\sqrt x + \sqrt a}<\frac{|x-a|}{\sqrt a}<\frac{\delta}{\sqrt a}$.

Now let's define $\delta = \varepsilon \sqrt a$.

Therefore: $|\sqrt x-\sqrt a|=\frac{|x-a|}{\sqrt x + \sqrt a}<\frac{|x-a|}{\sqrt a}<\frac{\delta}{\sqrt a}=\frac{\varepsilon \sqrt a}{\sqrt a}=\varepsilon$.

** Notice that $\sqrt x, \sqrt a>0$ so the expression $\sqrt x + \sqrt a $ doesn't need absolute value.

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