1.4 Continuous functions

(a) does not exist (b) $\infty$ (c) $\sqrt{a}$ (d) $-\sqrt{2}$ (e) $1$ (f) 0(g) does not exist

continuous

For $a \in (0, \infty)$ , show $\lim_{x \rightarrow a}\sqrt{x} = \sqrt{a}$ ．For all positive number $\varepsilon$ ，let $\displaystyle{\delta = \frac{\varepsilon}{\sqrt{a}}}$ ，then

$\displaystyle \vert\sqrt{x} - \sqrt{a}\vert = \frac{1}{\sqrt{x} + \sqrt{a}}\vert x - a\vert \leq \frac{\vert x - a\vert}{\sqrt{a}} \leq \varepsilon$

(a) $\max = 11, \min = -1$ (b) $\min = 1$

(c) $\displaystyle{\max = \left\{\begin{array}{cl} 4-2a & a \leq 0\\ 4-2a & 0 < a... ... -\frac{a^2}{4} & 0 < a < 4\\ -\frac{a^2}{4} & a \geq 4 \end{array} \right.}$

Put $\displaystyle{f(x) = 2\sin{x} - x}$ . Then $f(x)$ is continous at $\displaystyle{[\frac{\pi}{2},\pi]}$ and $\displaystyle{f(\frac{\pi}{2}) = 2 - \frac{\pi}{2} > 0}$ . Since $f(\pi) = - \pi < 0$ , by the intermediate value theorem， there exists $\xi$ in $\displaystyle{[(\frac{\pi}{2},\pi)}$ satisfying $f(\xi) = 0$ ．