Section 2.4 « Difference Equations to Differential Equations

Section 2.4

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1. (a) Since $f$ is a polynomial, $f$ is continuous at $t = 2$

(b) Since $f$ is a rational function, and is defined at $x = 17$ , $f$ is continuous at $x = 17$

(e) Since $h$ is not defined at $s = -1$ , $h$ is not continuous at $s = -1$

2. Since $\displaystyle{\lim_{t \to 2} g(t) = 7 = g(2)}$ , $g$ is continuous at $t = 2$ .

3. (a) $g$ is continuous on $(-\infty, \infty)$ .

(e) $f$ is continuous on $(-\infty, -2]$ and on $[2, \infty)$ .

4. $f$ is continuous on $(-\infty, 1)$ and on $[1, \infty)$ , but is not continuous at $x = 1$ .

5. $h$ is continuous on $(-\infty, -1]$ and on $(-1, \infty)$ , but $h$ is not continuous at $z = -1$ .

6. Yes, $f$ has a removable discontinuity at $t = 4$ . Let
$\displaystyle{g(t) = \begin{cases}\dfrac{t^2 - 7t + 12}{t - 4},& \text{if } t \ne 4,\\ 1,& \text{if } t = 4.\end{cases}}$

7. The discontinuity at $t = 5$ is not removable.

9. (a) $f$ is continuous on $(-\infty, \infty)$

(b) $g$ is continuous on $(-\infty, -1]$ , $(-1, 1)$ , and $[1, \infty)$ ; $g$ is not continuous at $x = -1$ and $x = 1$ .

(c) For $n = 0, \pm 1, \pm 2, \ldots$ , $h$ is continuous on intervals of the form $[2n, 2n + 1]$ and $(2n + 1, 2n)$ ; $h$ is not con.

11. For $n = 0, \pm 1, \pm 2, \ldots$ , $f$ is continuous on intervals of the form $((2n+1)\pi, 2n\pi)$ , $\Big[2n\pi, \big(2n + \frac{1}{2}\big)\pi\Big)$ , and $\Big(\big(2n + \frac{1}{2}\big)\pi, (2n + 1)\pi\Big]$ ; $f$ is not continuous at the points $x = \Big(2n + \frac{1}{2}\Big)\pi$ .

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Difference Equations to Differential Equations

Section 2.4

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