Solutions

(a) By definition,

$$\begin{aligned} \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N &\implies x_{n} < \epsilon^2 \\ &\implies x_{n} - \epsilon^2 < 0 \\ &\implies (\sqrt{x_{n}} - \epsilon)(\sqrt{x_{n}} + \epsilon ) < 0 \\ &\implies \sqrt{x_{n}} - \epsilon < 0 \\ &\implies \sqrt{x_{n}} < \epsilon. \end{aligned}$$

(b) By definition,

$$\begin{aligned} \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N &\implies \abs{x_{n} - x} < \epsilon\sqrt{x} \\ &\implies \abs{(\sqrt{x_{n}} - \sqrt{x})(\sqrt{x_{n}} + \sqrt{x})} < \epsilon\sqrt{x} \\ &\implies \abs{\sqrt{x_{n}} - \sqrt{x}} < \frac{\epsilon\sqrt{x}}{\sqrt{x_{n}} + \sqrt{x}} \leq \frac{\epsilon \sqrt{x}}{\sqrt{x}} = \epsilon. \end{aligned}$$

The last line is true because

$$\forall n \in \mathbb{N}, \quad \sqrt{x_{n}} \geq 0 \; \wedge \; x > 0.$$

The case where $x = 0$ is handled in (a).