Is this enough to satisfy continuity? Exercise:

Prove that the function $f(x) = \sqrt{x}$ is continuous for $x \geq 0$.


Attempt 1: original
First, I prove that the $\Delta f(x) \to 0$:
$\lim\limits_{\Delta x \to 0}{[f(x+\Delta x) - f(x)]}$
$ = \lim\limits_{\Delta x \to 0}{[\sqrt{x+\Delta x} - \sqrt{x}]}$
$ = \sqrt{x+0} - \sqrt{x}$
$ = 0$
Then, I prove that the limit of $f(x + \Delta x)$ as $\Delta x \to 0$ is equal to $f(x)$:
$\lim\limits_{\Delta x \to 0}{f(x + \Delta x)} =^? f(x)$
$\longrightarrow \lim\limits_{\Delta x \to 0}{\sqrt{x + \Delta x}} =^? \sqrt{x}$
$\longrightarrow \sqrt{x + 0} =^? \sqrt{x}$
$\longrightarrow \sqrt{x} = \sqrt{x}$
Is my attempt enough to prove the continuity of $f(x)$?

Attempt 2: after @ClementC.'s comment
$\lim\limits_{\Delta x \to 0}{[f(x+\Delta x) - f(x)]}$
$ = \lim\limits_{\Delta x \to 0}{[\sqrt{x+\Delta x} - \sqrt{x}]}$
$ = \lim\limits_{\Delta x \to 0}{[(\sqrt{x+\Delta x} - \sqrt{x})\frac{\sqrt{x+\Delta x} + \sqrt{x}}{\sqrt{x+\Delta x} + \sqrt{x}}]}$
$ = \lim\limits_{\Delta x \to 0}{[\frac{x+\Delta x - x}{\sqrt{x+\Delta x} + \sqrt{x}}]}$
$ = \lim\limits_{\Delta x \to 0}{[\frac{\Delta x}{\sqrt{x+\Delta x} + \sqrt{x}}]}$
But at this point, I'm left with the same dilemma: I can't replace $\Delta x$ with $0$, so what do I do?

I've been trying to use a similar method as the text in my book:

 A: You are using continuity of $f=\sqrt{}$ as soon as you write $\lim_{\Delta x\to 0} [\sqrt{x+\Delta x}-\sqrt{x}] = \sqrt{x+0}-\sqrt{x}$. So your proof is circular.
Moreover, it's not clear what the difference would be between your first and second step: for any fixed $x\geq 0$, proving $\lim_{\Delta x\to 0} [f(x+\Delta x)-f(x)] = 0$ is equivalent to proving $\lim_{\Delta x\to 0} f(x+\Delta x) = f(x)$.
This being said, let us give a proof of this continuity.

Fix any $x_0\geq 0$. For $h\in\mathbb{R}$ (for the sake of well-definedness, we assume $h\in[-x_0, \infty)$).


*

*If $x_0=0$:
$$\lvert f(x_0+h) - f(x_0)\rvert^2 = 
\sqrt{h}^2 = \lvert h\rvert \xrightarrow[h\to 0]{} 0
$$
so $f$ is continuous at $0$.

*If $x_0 > 0$: we have
$$\lvert f(x_0+h) - f(x_0)\rvert = \lvert\sqrt{x_0+h}-\sqrt{x_0}\rvert
= \sqrt{x_0}\cdot \left\lvert\sqrt{1+\frac{h}{x_0}}-1\right\rvert
$$
so it is sufficient to show that $f$ is continuous at $1$ (indeed, since $\frac{h}{x_0} \xrightarrow[h\to0]{}0$, this will imply that the second factor goes to $0$).
So let us do this:
$$
\lvert\sqrt{1+h}-1\rvert= \frac{\lvert \sqrt{1+h}^2-1^2\rvert}{\lvert\sqrt{1+h}+1\rvert}= \frac{\lvert h\rvert}{\lvert\sqrt{1+h}+1\rvert} \leq \lvert h\rvert\xrightarrow[h\to0]{}0
$$
proving continuity of $f$ at $1$, and therefore at any $x_0>0$ by the above discussion.
