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de2de:chapter-2 [2012/06/25 11:22]
dcs created
de2de:chapter-2 [2012/06/25 11:28] (current)
dcs
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 (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 $.
 +
 +----