Suppose we could prove the following lemma:
If $\lim_{h\rightarrow a} f(h) = m$ then
i) $\lim_{h\rightarrow a} (f(h) + w) = m + w = (\lim_{h\rightarrow a} f(h)) + w$.
ii) if $m \ge 0$ then $\lim_{h\rightarrow a}\sqrt{h} = \sqrt{m} = \sqrt{\lim_{h\rightarrow a}f(h)}$
iii) if $m \ne 0$ then $\lim_{h\rightarrow a}\frac{1}{f(h)} = 1/m = \frac 1{\lim_{h\rightarrow a}f(h)}$
Then we'd pretty much be done.
$\lim_{h\rightarrow 0}(x + h) = x + \lim_{h\rightarrow 0} h = x + 0=x$
$\lim_{h\rightarrow 0}\sqrt{x + h} = \sqrt{\lim_{h\rightarrow 0}(x+h)} = \sqrt{x}$
$\lim_{h\rightarrow 0}(\sqrt{x} + \sqrt{x+h}) = \sqrt{x} + \lim_{h\rightarrow 0}\sqrt{x + h} = \sqrt{x} + \sqrt{x} = 2\sqrt{x}$.
$\lim_{h\rightarrow 0}\frac 1{\sqrt{x} + \sqrt{x+h}} = \frac 1{\lim_{h\rightarrow 0}(\sqrt{x} + \sqrt{x+h})} = 1/2\sqrt{x}$.
Done.
Must prove lemmma.
i) For $\epsilon > 0$ there is a $\delta$ so that $|h - a| < \delta \implies |f(h) - m| < \epsilon$ so $|h-a| < \delta \implies |(f(h)+w)-(m+w)|= |f(h) -m| < \epsilon$.
ii) For $2\sqrt{m} > \epsilon > 0$ let $\gamma = 2\sqrt{m}\epsilon - \epsilon^2 > 0$.
There is a $\delta$ so that $|h - a| < \delta \implies |f(h) - m| < \gamma$
$\implies m - \gamma < f(h) < m + \gamma $
$\implies (\sqrt{m} - \epsilon)^2 = m - 2\sqrt{m}\epsilon + \epsilon^2 < f(h) < m + 2\sqrt{m}\epsilon - \epsilon^2 < m + 2\sqrt{m}\epsilon + \epsilon^2=(\sqrt{m} + \epsilon)^2$
$\implies \sqrt{m} - \epsilon < f(h) < \sqrt{m} + \epsilon$
$\implies |f(h) - \sqrt{m}| < \epsilon$.
iii) For $ \epsilon > 0$. Let $\epsilon_2 = \min(|m|/2, m^2*\epsilon/2) > 0$. (Remember $m \ne 0$, so we can be assured $\epsilon_2 > 0$.)
We can find a $\delta$ so that $|h-a| < \delta \implies |f(h) - m| < \epsilon_2$.
So $|h-a| < \delta \implies |f(h) - m| < \epsilon_2 \le |m|/2 \implies |f(h)| > |m|/2$.
So if $|h - a| < \delta$ then
$|\frac 1{f(h)} - \frac 1m| = |\frac{m - f(h)}{m*f(h)}| = |m - f(h)|*\frac 1{|m*f(h)|}$
$< |m-f(h)|*\frac 1{m^2/2} < \epsilon_2*\frac 2{m^2}$
$\le \frac{m^2\epsilon}{2}*\frac 2{m^2}=\epsilon$.
So we proved the lemma and thus the result.
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Actually let's see what Mr. Rudin says. In Principles of Mathematical Analysis by walter rudin we have in chapter 4.
Theorem 4.4. Suppose $E \subset X$ a metric space (we'll just say $X$ and $E$ are $\mathbb R$), $p$ is a limit point of $E$ (that just means we talk about $x \rightarrow p$), $f$ and $g$ are complex functions on $E$ (let's just say $f$ and $g$ are real functions) and
$\lim_{x \rightarrow p}f(x) = A$, $\lim_{x\rightarrow p}g(x) = B$ then
a) $\lim_{x \rightarrow p}(f + g)(x)=A+B$
b) $\lim_{x\rightarrow p}(fg)(x) = AB$
c) $\lim_{x\rightarrow p}(f/g)(x) = A/B$ if $B\ne 0$.
The proof of a) and c) are as I gave them. b) is pretty basic and similar.
Then Def 4.5 defines continuous functions. ($f$ is continuous at $p$, if for every $\epsilon > 0$ there exists a $\delta > 0$ so that for all $x$ where $d(x,p)< \delta$ it follows that $d(f(x),f(p)) < \epsilon$)
Theorem 4.6: $f$ is continuous at $p$ if and only if $\lim_{x\rightarrow p}f(x) = f(p)$. This follows purely by definitions.
Thereom 4.7: says that composition of continuous functions are continuous. The proof is simple. Basically you find $\epsilon, \gamma, \delta$ so that $|x-p|< \gamma \implies |f(x) - f(p)| < \epsilon$ and $|x-p|<\delta \implies |h(x) - h(p)|<\gamma$ there for $|x-p| < \delta \implies |h(x) - h(p)| < \gamma \implies |f(h(x)) - f(h(p))| < \epsilon$ so $f(h(x))$ is continuous at $p$..
With those we just have to show $\sqrt{x}$ is continuous which is basically my lemma ii).
