1.4 Continuous functions

1.

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

2.

continuous

3.

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

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

4.

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

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

5.

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