1.5 Answer

1.5

1.

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

Also，$f(2) = 3$. Then $f(x)$ is continuous at $x = 2$.

2.

For $a \in (0, \infty)$, we show $\lim_{x \rightarrow a}\sqrt{x} = \sqrt{a}$．For any $\varepsilon$，we set $${\delta = \frac{\varepsilon}{\sqrt{a}}}$$. Then

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

3.

(a) $${f(x) = x^2 - 3x + 1 = (x - \frac{3}{2})^2 - \frac{5}{4}}$$. 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) $${\max = \left\{\begin{array}{cl} 4-2a & a \leq 0\\ 4-2a & 0 < a... ... -\frac{a^2}{4} & 0 < a < 4\\ -\frac{a^2}{4} & a \geq 4 \end{array} \right.}$$

4.

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