Show that if $(x_n) \rightarrow x$ then $(\sqrt{x_n}) \rightarrow \sqrt{x}$. Let $x_n \ge 0$ for all $n \in \mathbf{N}$ and $x>0$, show that if $(x_n) \rightarrow x$ then $(\sqrt{x_n}) \rightarrow \sqrt{x}$.
My textbook does the following proof:
Let $\epsilon >0$, we must find an $N$ such that $n \ge N$ implies $|\sqrt{x_n} - \sqrt{x}|< \epsilon$ for all $n \ge N$. 
\begin{align}
|\sqrt{x_n} - \sqrt{x}| &= |\sqrt{x_n} - \sqrt{x}|\left(\frac{\sqrt{x_n} + \sqrt{x}}{\sqrt{x_n} + \sqrt{x}}\right) \\
& = \frac{|x_n - x|}{\sqrt{x_n} + \sqrt{x}} \\
& \le \frac{|x_n - x|}{\sqrt{x}} \ \ \ \ \ \ \cdots \ \ \ \ \ \ (1)
\end{align}
Since $(x_n) \rightarrow x$ and $x>0$, we can choose $N$ such that $|x_n - x| < \epsilon\sqrt{x}$ whenever $n \ge N$ and so for all $n \ge N$, $|\sqrt{x_n} - \sqrt{x}| < \frac{\epsilon \sqrt{x}}{\sqrt{x}} = \epsilon$.
What I'm wondering is, in Eqn.$(1)$, why is $\sqrt{x}$ kept in the denominator? Couldn't one just have simply $\le |x_n - x|$ and choose $N$ such that $|x_n - x| < \epsilon$ for $n \ge N$ and the rest follows?

Because $\sqrt{x_n}+\sqrt{x}\geq \sqrt x$ and thus $$\frac{1}{\sqrt{x_n}+\sqrt x}\leq \frac{1}{\sqrt x}.$$
When you'll see continuous function, such a proof is easier using the continuity of $x\longmapsto \sqrt x$.
