If $(x_{n}) \rightarrow x$, show that $\sqrt{x_{n}} \to \sqrt{x}$

Question

If $(x_{n}) \rightarrow x$, show that $\sqrt{x_{n}} \rightarrow \sqrt{x}$ for $x > 0$.

Let $\epsilon > 0$ be arbitrary, want to find $N \in \mathbb{N}$ such that $n \geq N \Rightarrow |\sqrt{x_{n}} - \sqrt{x}| < \epsilon$.

$$|\sqrt{x_{n}} - \sqrt{x}| = | \sqrt{x_{n}} - \sqrt{x}| \cdot \left(\frac{\sqrt{x_{n}} + \sqrt{x} }{\sqrt{x_{n}} + \sqrt{x}}\right) = \frac{|x_{n} - x|}{\sqrt{x_{n}} + \sqrt{x}} \leq \frac{|x_{n} - x|}{\sqrt{x}} < \epsilon.$$

So choose $N$ such that $|x_{n} - x| < \epsilon \sqrt{x}$.

Is this correct?

Answer 1

As mentioned in the comments, your proof is correct. Some people would argue that the line

Choose $N$ such that $\vert x_n-x\vert <\epsilon\sqrt x$.

requires some extra explanation on why such an $N$ exists. This is of course due to the fact that $(x_n)\to x$.