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1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٢ صفحة 2. –/1 pointsSEssCalcET2 1.3.005. For the function g whose graph is given, state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.) (a) (b) (c) (d) (e) (f) (g) (h) lim g(t) t → 0− lim g(t) t → 0+ lim g(t) t → 0 lim g(t) t → 2− lim g(t) t → 2+ lim g(t) t → 2 g(2) lim g(t) t → 4 1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٣ صفحة 3. –/1 pointsSEssCalcET2 1.3.007. Sketch the graph of an example of a function f that satisfies all of the given conditions. lim f(x) = 1, f(x) = 2, f(0) = −1 x → 0− lim x → 0+ 1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٤ صفحة 4. –/1 pointsSEssCalcET2 1.3.023. Use the given graph of to find a number δ such that δ = 5. –/1 pointsSEssCalcET2 1.3.033. Prove the statement using the ε, δ definition of a limit. Given ε > 0, we need δ —Select— such that if 0 < |x − 1| < δ, then But So if we choose then Thus, by the definition of a limit. 6. –/1 pointsSEssCalcET2 1.3.035. Prove the statement using the ε, δ definition of a limit. Given we need such that if then 㱻 㱻 By the definition of a limit, f(x) = x if |x − 4| < δ then x − 2 < 0.4. lim = 4 x → 1 13 + 3x 4 − 4 —Select— . 13 + 3x 4 − 4 < ε 㱻 < ε 㱻 |x − 1| < ε 㱻 |x − 1| < —Select— . 13 + 3x 4 3x − 3 4 3 4 δ = —Select— , 0 < |x − 1| < δ − 4 0, δ —Select— 0 < |x − 3| < δ, − 7 —Select— 㱻 x2 + x − 12 x − 3 − 7 < ε (x + 4)(x − 3) x − 3 |x + 4 − 7| < ε [x ≠ 3] 0 < |x − 3| < ε. So choose δ = —Select— . Then 0 < |x − 3| < δ 0 < |x − 3| < ε 0 < |x + 4 − 7| < ε − 7 < ε [x ≠ 3] (x + 4)(x − 3) x − 3 − 7 0, we need δ —Select— such that if then or upon simplifying we need whenever Notice that if So take Then and so Thus, by the definition of a limit, lim f(x) = ∞ x → −8 f(−8) = ∞ lim f(x) = −∞ x → 5+ f(5) = −∞ (x lim 2 − 1) = 8 x → −3 0 < |x − (−3)| < δ, |(x2 − 1) − 8| —Select— |x2 − 9| < ε 0 < |x + 3| < δ. |x + 3| < 1, then −1 < x + 3 < 1 −7 < x − 3 < −5 |x − 3| < —Select— . δ = —Select— . 0 < |x + 3| < δ |x − 3| < 7 |x + 3| < ε/7, |(x2 − 1) − 8| = |(x + 3)(x − 3)| = |x + 3||x − 3| < (ε/7)(7) = —Select— . (x lim 2 − 1) = . x → −3 1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٦ صفحة 9. –/1 pointsSEssCalcET2 1.3.533.XP.MI. A tank holds 3000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes. t (min) 5 10 15 20 25 30 V (gal) 2046 1332 780 324 84 0 (a) If P is the point (15, 780) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph with the following values. (Round your answers to one decimal place.) Q slope (5, 2046) (10, 1332) (20, 324) (25, 84) (30, 0) (b) Estimate the slope of the tangent line at P by averaging the slopes of two adjacent secant lines. (Round your answer to one decimal place.) 10.–/1 pointsSEssCalcET2 1.3.AE.006. Video Example EXAMPLE 6 The Heaviside function H is defined by [This function is named after the electrical engineer Oliver Heaviside (1850- 1925) and can be used to describe an electric current that is switched on at time t = 0.] Its graph is shown in the figure. As t approaches 0 from the left, H(t) approaches . As t approaches 0 from the right, H(t) approaches . Therefore the limit as t approaches 0 of H(t) does not exist. H(t) = 0 if t 0, we need δ > 0 such that if then Take Then Thus, by the definition of a limit. x lim 2 = 0 x → 0 0 < |x − 0| < δ, |x2 − 0| < ε 㱻 < ε 㱻 |x| < . δ = . 0 < |x − 0| < δ |x2 − 0| < ε. x lim 2 = 0 x → 0

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1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٢ صفحة 2. –/1 pointsSEssCalcET2 1.3.005. For the function g whose graph is given, state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.) (a) (b) (c) (d) (e) (f) (g) (h) lim g(t) t → 0− lim g(t) t → 0+ lim g(t) t → 0 lim g(t) t → 2− lim g(t) t → 2+ lim g(t) t → 2 g(2) lim g(t) t → 4 1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٣ صفحة 3. –/1 pointsSEssCalcET2 1.3.007. Sketch the graph of an example of a function f that satisfies all of the given conditions. lim f(x) = 1, f(x) = 2, f(0) = −1 x → 0− lim x → 0+ 1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٤ صفحة 4. –/1 pointsSEssCalcET2 1.3.023. Use the given graph of to find a number δ such that δ = 5. –/1 pointsSEssCalcET2 1.3.033. Prove the statement using the ε, δ definition of a limit. Given ε > 0, we need δ —Select— such that if 0 < |x − 1| < δ, then But So if we choose then Thus, by the definition of a limit. 6. –/1 pointsSEssCalcET2 1.3.035. Prove the statement using the ε, δ definition of a limit. Given we need such that if then 㱻 㱻 By the definition of a limit, f(x) = x if |x − 4| < δ then x − 2 < 0.4. lim = 4 x → 1 13 + 3x 4 − 4 —Select— . 13 + 3x 4 − 4 < ε 㱻 < ε 㱻 |x − 1| < ε 㱻 |x − 1| < —Select— . 13 + 3x 4 3x − 3 4 3 4 δ = —Select— , 0 < |x − 1| < δ − 4 < ε. 13 + 3x 4 lim = 4 x → 1 13 + 3x 4 lim = 7 x → 3 x2 + x − 12 x − 3 ε > 0, δ —Select— 0 < |x − 3| < δ, − 7 —Select— 㱻 x2 + x − 12 x − 3 − 7 < ε (x + 4)(x − 3) x − 3 |x + 4 − 7| < ε [x ≠ 3] 0 < |x − 3| < ε. So choose δ = —Select— . Then 0 < |x − 3| < δ 0 < |x − 3| < ε 0 < |x + 4 − 7| < ε − 7 < ε [x ≠ 3] (x + 4)(x − 3) x − 3 − 7 < ε. x2 + x − 12 x − 3 lim = . x → 3 x2 + x − 12 x − 3 1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٥ صفحة 7. –/1 pointsSEssCalcET2 1.3.503.XP. Explain the meaning of each of the following. (a) The values of f(x) can be made arbitrarily large by taking x sufficiently close to (but not equal to) −8. The values of f(x) can be made arbitrarily close to −8 by taking x sufficiently large. The values of f(x) can be made arbitrarily close to 0 by taking x sufficiently close to (but not equal to) −8. (b) The values of f(x) can be made arbitrarily close to −∞ by taking x sufficiently close to 5. The values of f(x) can be made negative with arbitrarily large absolute values by taking x sufficiently close to, but greater than, 5. As x approaches 5, f(x) approaches −∞. 8. –/1 pointsSEssCalcET2 1.3.531.XP. Prove the statement using the ε, δ definition of a limit. Given ε > 0, we need δ —Select— such that if then or upon simplifying we need whenever Notice that if So take Then and so Thus, by the definition of a limit, lim f(x) = ∞ x → −8 f(−8) = ∞ lim f(x) = −∞ x → 5+ f(5) = −∞ (x lim 2 − 1) = 8 x → −3 0 < |x − (−3)| < δ, |(x2 − 1) − 8| —Select— |x2 − 9| < ε 0 < |x + 3| < δ. |x + 3| < 1, then −1 < x + 3 < 1 −7 < x − 3 < −5 |x − 3| < —Select— . δ = —Select— . 0 < |x + 3| < δ |x − 3| < 7 |x + 3| < ε/7, |(x2 − 1) − 8| = |(x + 3)(x − 3)| = |x + 3||x − 3| < (ε/7)(7) = —Select— . (x lim 2 − 1) = . x → −3 1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٦ صفحة 9. –/1 pointsSEssCalcET2 1.3.533.XP.MI. A tank holds 3000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes. t (min) 5 10 15 20 25 30 V (gal) 2046 1332 780 324 84 0 (a) If P is the point (15, 780) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph with the following values. (Round your answers to one decimal place.) Q slope (5, 2046) (10, 1332) (20, 324) (25, 84) (30, 0) (b) Estimate the slope of the tangent line at P by averaging the slopes of two adjacent secant lines. (Round your answer to one decimal place.) 10.–/1 pointsSEssCalcET2 1.3.AE.006. Video Example EXAMPLE 6 The Heaviside function H is defined by [This function is named after the electrical engineer Oliver Heaviside (1850- 1925) and can be used to describe an electric current that is switched on at time t = 0.] Its graph is shown in the figure. As t approaches 0 from the left, H(t) approaches . As t approaches 0 from the right, H(t) approaches . Therefore the limit as t approaches 0 of H(t) does not exist. H(t) = 0 if t < 0 1 if t ≥ 0. 1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٧ صفحة 11.–/1 pointsSEssCalcET2 1.3.AE.007. Video Example EXAMPLE 7 The graph of a function g is shown in the figure. Use it to state the values (if they exist) of the following: SOLUTION From the graph we see that the values of g(x) approach as x approaches 2 from the left, but they approach as x approaches 2 from the right. Therefore (c) Since the left and right limits are different, we conclude that the limit as x approaches 2 of g(x) does not exist. The graph also shows that (f) This time, the left and right limits are the same and so, by this theorem, we have Despite this fact, notice that g(5) ≠ 1. (a) lim g(x) x → 2− (b) lim g(x) x → 2+ (c) lim g(x) x → 2 (d) lim g(x) x → 5− (e) lim g(x) x → 5+ (f) lim g(x). x → 5 (a) lim g(x) = and (b) g(x) = . x → 2− lim x → 2+ (d) lim g(x) = and (e) g(x) = . x → 5− lim x → 5+ lim g(x) = x → 5 1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٨ صفحة 12.–/1 pointsSEssCalcET2 1.3.AE.005. EXAMPLE 4 Investigate the following limit. SOLUTION Again the function is undefined at 0. Evaluating the function for some small values of x, we get Similarly, On the basis of this information we might be tempted to guess that but this time our guess is wrong. Note that although for any integer n, it is also true that f(x) = 1 for infinitely many values of x that approach 0. You can see this from the graph of f given in the figure. The compressed lines near the y-axis indicate that the values of f(x) oscillate between 1 and −1 infinitely often as x approaches 0. (Use a graphing device to graph f(x) and zoom in toward the origin several times. What do you observe?) Since the values of f(x) do not approach a fixed number as x approaches 0, does not exist. 13.–/1 pointsSEssCalcET2 1.3.015. Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically. (Round your answer to two decimal places.) 14.–/1 pointsSEssCalcET2 1.3.017. Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically. (Round your answer to two decimal places.) lim sin x → 0 π x f(x) = sin(π/x) f(1) = sin π = f = sin 2π = 0 f = sin 3π = f = sin π = 0 f(0.1) = sin 10π = 0 f(0.01) = sin 100π = . 1 2 1 3 1 4 f(0.001) = f(0.0001) = 0. lim sin = , x → 0 π x f(1/n) = sin nπ = lim sin x → 0 π x lim x → 0 − 9 x x + 81 lim x → 1 x9 − 1 x4 − 1 1.3 Homework ! ٨:١٧ ،٢٠١٧/٨/٢٥ https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ٩ من ٩ صفحة 15.–/1 pointsSEssCalcET2 1.3.039. Prove the statement using the ε, δ definition of a limit. Given ε > 0, we need δ > 0 such that if then Take Then Thus, by the definition of a limit. x lim 2 = 0 x → 0 0 < |x − 0| < δ, |x2 − 0| < ε 㱻 < ε 㱻 |x| < . δ = . 0 < |x − 0| < δ |x2 − 0| < ε. x lim 2 = 0 x → 0

 
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