Let $f:[0,\infty)\to \mathbb{R}$ be given by $f(x)=\sqrt{x}$. Show that it is continuous.
This is taken from Example 3.7 on <link> page 22 on the paper. It has shown that it is continuous at $c=0$ and $c>0$ seperately, which I understood why. I wondered how I would prove it if I got a similar problem from my teacher without showing how I have processed it (that is, how the values of $\delta>0$ were found). I thought about the following way,
Let $\epsilon>0$ be given and let $c\in [0,\infty)$. Choose $$\delta=\begin{cases} \epsilon^{2} & \text{ if } c=0 \\ \epsilon\sqrt{c} & \text{ if } c>0. \end{cases}$$ Assume $c=0$. If $\left | x-c \right |=\left | x-0 \right |<\delta$, then $$\left | f(x)-f(c) \right |=\left | \sqrt{x}-0 \right |=\left | x-0 \right |^{1/2}<\delta^{1/2}=\epsilon.$$ Assume $c>0$. If $\left | x-c \right |<\delta$, then $$\left | f(x)-f(c) \right |= \left | \sqrt{x}-\sqrt{c} \right |=\left | \frac{x-c}{\sqrt{x}+\sqrt{c}} \right |\leq \frac{1}{\sqrt{c}}\left | x-c \right | <\frac{1}{\sqrt{c}}\delta=\epsilon.$$ This proves that $f$ is continuous.
Is this correct, or is there a better way to show it mathematically?
Looks good! About the only tweak that I would suggest is to make use of the fact that $x\mapsto\sqrt x$ is a monotonically increasing function on the nonnegative reals. Do you see why that is relevant in both cases?