1.5 Answer

1.5

$\displaystyle lim_{x \to 2}\frac{x^2 - x - 2}{x - 2} = \lim_{x \to 2}\frac{(x+1)(x-2)}{x-2} = 3$

Also，

. Then

is continuous at $x = 2$

For $a \in (0, \infty)$ , we show $\lim_{x \rightarrow a}\sqrt{x} = \sqrt{a}$ ．For any $\varepsilon$ ，we set $\displaystyle{\delta = \frac{\varepsilon}{\sqrt{a}}}$ . 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) $\displaystyle{f(x) = x^2 - 3x + 1 = (x - \frac{3}{2})^2 - \frac{5}{4}}$ . Then the maximum is at $x = -2$ and $\max = 11$ . Also, the minimum is at $x = 1$ and $\min = -1$

(b) As $x \to 0+$ , $f(x) = \frac{1}{x}$ approaches to $\infty$ . Then no maximum and $\min = 1$

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

Let $\displaystyle{f(x) = 2\sin{x} - x}$ . Then $f(x)$ is continuous on $\displaystyle{[\frac{\pi}{2},\pi]}$ . Since $\displaystyle{f(\frac{\pi}{2}) = 2 - \frac{\pi}{2} > 0}$ and $f(\pi) = - \pi < 0$ . Thus by the mean-value theorem, we have $\xi$ in $\displaystyle{[(\frac{\pi}{2},\pi)}$ so that $f(\xi) = 0$ .