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 de2de:chapter-2 [2012/06/25 11:22]dcs created de2de:chapter-2 [2012/06/25 11:28] (current)dcs 2012/06/25 11:28 dcs 2012/06/25 11:22 dcs created 2012/06/25 11:28 dcs 2012/06/25 11:22 dcs created Line 114: Line 114: (e) $\displaystyle{\lim_{x \to 0^+}\lfloor \cos(x) \rfloor = 0}$ (e) $\displaystyle{\lim_{x \to 0^+}\lfloor \cos(x) \rfloor = 0}$ + + 5. (b) $\displaystyle{\lim_{z \to 2^-}g(z) = 5}$ + + %%(c)%% $\displaystyle{\lim_{z \to 2^+}g(z) = 5}$ + + (d) Yes: $\displaystyle{\lim_{z \to 2}g(z) = 5}$ + + 7. (a) $\displaystyle{\lim_{x \to \infty} (3x + 4) = \infty}$ + + (b) $\displaystyle{\lim_{x \to \infty} \frac{x^3 + 3x - 1}{2x^3 - x^2 + 21} = \frac{1}{2}}$ ​ + + %%(c)%% $\displaystyle{\lim_{u \to \infty} \frac{u^4 + 3u - 6}{3u^2 + 1} = \infty}$ ​ + + (e) $\displaystyle{\lim_{x \to -\infty} \frac{x^5 - 6x + 13}{x^2 + 18x - 25} = -\infty}$ ​ + + (g) $\displaystyle{\lim_{v \to \infty} \sqrt{\frac{2v + 1}{v - 2}} = \sqrt{2}}$ ​ + + (i) $\displaystyle{\lim_{x \to \infty} \frac{3x + 1}{\sqrt{4x^2 + 5}} = \frac{3}{2}}$ ​ + + (j) $\displaystyle{\lim_{x \to -\infty} \frac{3x + 1}{\sqrt{4x^2 + 5}} = -\frac{3}{2}}$ + + 9. $\displaystyle{\lim_{x \to \frac{\pi}{2}^-} \tan(x) = \infty}$, $\displaystyle{\lim_{x \to \frac{\pi}{2}^+} \tan(x) = -\infty}$, $\displaystyle{\lim_{x \to \frac{\pi}{2}} \tan(x)}$ does not exist + + 11. (a) Use the fact that $-1 \le \sin(x) \le 1$ for all $x > 0$. + + (b) $\displaystyle{\lim_{x \to \infty}\frac{\sin(x)}{x} = 0}$ + + ---- + + **Section 2.4** + + 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 $x = 17$ + + %%(c)%% Since $f$ is not defined at $x = 16$, $f$ is not continuous at $x = 16$ + + (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 $(-\infty, \infty)$. + + %%(c)%% $g$ is continuous on $(-\infty, 0)$ and on $(0, \infty)$. + + (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 + $\displaystyle{g(t) = \begin{cases}\displaystyle{\frac{t^2 - 7t + 12}{t - 4}},& \text{if } t \ne 4,\\ 1,& \text{if } t = 4.\end{cases}}$ ​ + + 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 continuous at $t = 0, \pm 1, \pm 2, \ldots$. ​ + + 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 $x = \Big(2n + \frac{1}{2}\Big)\pi$. + + ---- + + **Section 2.5** + + 1. (a) $1.418$ + + %%(c)%% $0.739$ + + 2. (b) $2$ + + %%(c)%% $-0.739$ and $0.739$ + + 3. (b) $\$51.28 $+ + %%(c)%%$ 979 $units + + (d) If the producers raised the price above the equilibrium price, then demand would decrease while supply increased. With supply exceeding demand, the producers would eventually decrease the price. + If the producers lowered the price below the equilibrium price, then demand would increase while supply decreased. With the demand exceeding the supply, the producers would eventually increase the price. + + (e) If the producers increased production, the supply curve would be higher and the equilibrium price would decrease. + If the producers decreased production, the supply curve would be lower and the equilibrium price would increase. + + + 4.$ 125 \text{ yards } \times 250 \text{ yards} $+ + + 5.$\$10.00$ + + 6. (a) No + + (b) Yes, $-1$ is the minimum value of $h$ on $[-1, 2)$. + + %%(c)%% Yes, because the Extreme Value Theorem does not apply in this case  since the interval $[-1, 2)$ is not closed. + + + 7. (a) No + + (b) Yes, because the Intermediate Value Theorem does not apply in this case since the Heaviside function is not continuous on $[-1, 1]$. + + %%(c)%% $H$ has a maximum value of $1$ and a minimum value of $-1$ on $[-1, 1]$. + + (d) Yes; although the Extreme Value Theorem does not apply in this case (since $H$ is not continuous on $[-1, 1]$), that does not preclude $H$ from having maximum and minimum values on the interval. ​ + + + 9. Apply the Intermediate Value Theorem to the function $h(x) = f(x) - g(x)$. + + 10. Apply the Intermediate Value Theorem to the function $g(x) = f(x) - x$. + + ----

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