1.3 Homework ! : // 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) lim g(t) t 0 t 0+ lim g(t) t0 lim g(t) lim g(t) t 2 t 2+ lim g(t) t2 g(2) lim g(t) t4 https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1.3 Homework ! : // 3. -/1 pointsSEssCalcET2 1.3.007. Sketch the graph of an example of a function f that satisfies all of the given conditions. lim x 0 f(x) = 1, lim x 0+ f(x) = 2, f(0) = 1 https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1.3 Homework ! : // 4. -/1 pointsSEssCalcET2 1.3.023. x to find a number such that Use the given graph of f(x) = if |x 4| < then x 2 < 0.4. = 5. -/1 pointsSEssCalcET2 1.3.033. Prove the statement using the , definition of a limit. lim x1 13 + 3x =4 4 Given > 0, we need 13 + 3x 4 < 4 0 < |x 1| < ---Select--- 13 + 3x 4 4 such that if 0 < |x 1| < , then ---Select--- . But 3x 3 3 < |x 1| < |x 1| < ---Select--. So if we choose = 4 4 13 + 3x 13 + 3x 4 < . Thus, lim = 4 by the definition of a limit. 4 4 x1 ---Select--- , then 6. -/1 pointsSEssCalcET2 1.3.035. Prove the statement using the , definition of a limit. lim x3 x2 + x 12 = 7 x3 Given > 0, we need (x + 4)(x 3) 7 < x3 Then 0 < |x 3| < ---Select--- such that if 0 < |x 3| < , then |x + 4 7| < [x 3] 0 < |x 3| < x2 + x 12 7 x3 0 < |x 3| < . So choose = 0 < |x + 4 7| < ---Select--- . (x + 4)(x 3) 7 < [x 3] x3 x2 + x 12 7 < . By the definition of a limit, lim x2 + x 12 = x3 x3 x3 https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 ---Select--- . 1.3 Homework ! : // 7. -/1 pointsSEssCalcET2 1.3.503.XP. Explain the meaning of each of the following. (a) lim x 8 f(x) = 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. f(8) = The values of f(x) can be made arbitrarily close to 0 by taking x sufficiently close to (but not equal to) 8. (b) lim x 5+ f(x) = 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. f(5) = As x approaches 5, f(x) approaches . 8. -/1 pointsSEssCalcET2 1.3.531.XP. Prove the statement using the , definition of a limit. lim x 3 (x2 1) = 8 Given > 0, we need ---Select--- such that if 0 < |x (3)| < , then |(x2 1) 8| ---Select--- we need |x2 9| < whenever 0 < |x + 3| < . Notice that if |x + 3| < 1, then 1 < x + 3 < 1 |x 3| < ---Select--- . So take = ---Select--- . Then 0 < |x + 3| < |(x2 1) 8| = |(x + 3)(x 3)| = |x + 3||x 3| < (/7)(7) = lim x 3 (x2 1) = ---Select--- or upon simplifying 7 < x 3 < 5 |x 3| < 7 and |x + 3| < /7, so . Thus, by the definition of a limit, . https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1.3 Homework ! : // 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) V (gal) 5 10 15 20 25 30 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. EXAMPLE 6 The Heaviside function H is defined by H(t) = 0 1 if t < 0 if t 0. [This function is named after the electrical engineer Oliver Heaviside (18501925) and can be used to describe an electric current that is switched on at time t = 0.] Its graph is shown in the figure. Video Example As t approaches 0 from the left, H(t) approaches approaches 0 from the right, H(t) approaches . As t . Therefore the limit as t approaches 0 of H(t) does not exist. https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1.3 Homework ! : // 11.-/1 pointsSEssCalcET2 1.3.AE.007. 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: (a) (d) lim g(x) (b) lim g(x) (e) x 2 x 5 SOLUTION lim g(x) (c) lim g(x) (f) x 2+ x 5+ lim g(x) x2 lim g(x). x5 From the graph we see that the values of g(x) approach x approaches 2 from the left, but they approach as as x approaches 2 from the right. Therefore Video Example (a) lim x 2 g(x) = and (b) lim x 2+ g(x) = . (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 (d) lim x 5 g(x) = and (e) lim x 5+ g(x) = . (f) This time, the left and right limits are the same and so, by this theorem, we have lim g(x) = x5 Despite this fact, notice that g(5) 1. https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1.3 Homework ! : // 12.-/1 pointsSEssCalcET2 1.3.AE.005. EXAMPLE 4 Investigate the following limit. lim sin x0 x Again the function f(x) = sin(/x) is undefined at 0. Evaluating SOLUTION the function for some small values of x, we get 1 2 1 f 4 f f(1) = sin = f 1 3 = sin 3 = f(0.1) = sin 10 = 0 = sin 2 = 0 = sin =0 f(0.01) = sin 100 = . Similarly, f(0.001) = f(0.0001) = 0. On the basis of this information we might be tempted to guess that lim sin x0 = x , but this time our guess is wrong. Note that although f(1/n) = sin n = 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, lim sin x0 does not exist. x 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.) lim x0 x + 81 9 x 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 x9 1 x 1 x4 1 https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1.3 Homework ! : // 15.-/1 pointsSEssCalcET2 1.3.039. Prove the statement using the , definition of a limit. lim x2 = 0 x0 Given > 0, we need > 0 such that if 0 < |x 0| < , then |x2 0| < |x| < . < = Take . Then 0 < |x 0| < |x2 0| < . Thus, lim x2 = 0 by the x0 definition of a limit. https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 Nee-mm? EXAMPLE 7 The graph ofa function g is shown in the gure. Use itto state the values (if they exist) of the following: (a) Hm gix) (b) lim+gm (Cixlijngm XIZT xIZ (cl) lim QIX) (e) lim 51x) (f) lim g(x). xa-S'l' x35 xa-S' SOLUTION From the graph we see that the values ofg(x) approach D as x approaches 2 from the left, but they approach |:| asx approaches 2 from the right. Therefore (a) lim g(x)=|_3__l and (b) lim g(x}=|E|. X3-2_ X3-2+ (c) Since the left and right limits are different, we conclude that the limit as x approaches 2 of gix) does not exist. The graph also shows that (d) lim5_g(x) = and (e) lim5+g(x) = E . xs xs (f) Thistime, the left and right limits are the same and so, by this theorem, we have lim 2 H 59m Iii Despite thisfact, notice that 9(5) at 1. BCAMPLE 4 Invesb'gate the following limit. lim 9n E X3 0 X SOLUTION Again the Jncb'on x) 2 sin(m'x) is undened at D. Evaluating the function for some small values ofx, we get _ - _ i _ - _ llisinErIEI (2)sm27r () sin33r=|E| F(%)=sin|II3r=D 110.1) = sin 103: 0 110.01) sin 10cm: E. Similarly, 110.001) = 110.0001) = D. On the bags ofdris information we might be tempted to guessthat . . 9r an a = E . but thisb'me our guess is wrong. Note that although lm) = sin nfr = El for any integer n, it is also true mat for) = 1 for infinitely many values ofx that approach D. You can see misfrom the graph of 1' given in the gure. The compressed lines near the yaxis indicate that the values of for) oscillate between 1 and 1 innitely often asx approaches D. (Use a graphing device to graph for) and zoom in toward the origin several times. iil'vhat do you observe?) Since the values of for) do not approach a xed number asx approaches D, lim sinE does not exist. XhO X 13.,.I'l poir'ItSSESSCalcEI'Q 1.3.015. Use a table of valuesoo estimate the value ofthe limit. If you have a graphing device, use it to conrm your result graphically. (Round your answer to two decimal places.) Iil'l] VX+819 XZIU X III Need Help? 14. ,.I'l poir'ItSSESSCalcEI'Q 1.3.011 Use a table of valuesto estimate the value ofthe limit. If you have a graphing device, use it to conrm your result graphically. (Round your answer to two decimal places.) 9_ lim X 1 Xslel IE needHenp? \f\f\f7 + Automatic Zoom 4. f1 poinisSEssCalcEm 1 .3 .023 . Use the given graph of for) : J; to nd a number 6 such that 5. -f1 pounISSEssCalcElz 1.1033. Prove the statement using the a, 5 denition of a limit. 13+3x lim = 4 x-:l Gvene >0, we need 5 >0\" asuchthatifo < |x 1| < 6, then 13+3x_ 4 |#4| <5 3 But 4|<|3X43|(FEMI. Atank holds 3000 gallons ofwater, which drains from the bottom of the tank in half an hour. The values in the table show the volume Vof water remaining in the tank (in gallons) after tminutes. ['(min) 5 ll] 15 2D 25 30 V(ga|) 2046 1332 780 324 84 o (a) HP is the point(15, ?80) on the graph of If, find the slopes ofthe secant linequ when Q isthe point on the graph with the following values. (Round your answersto one decimal place.) Q slope (5, 2046: m (10, 1332) m (20. 324: El (25. 84) [EE- (sn. 0) El (b) Estimate the slope ofthe tangent line at P by averaging the slopes oftwo adjacent secant lines. (Round your answer to one decimal place.) m Need Help? ii : 10.1'1 pointsSECalcEl'Z 1.3.AE.006. EXAMPLE 6 The Heaviside function H is defined by _ 0 if t < 0 Hal{1 iftz 0. [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 gure. Video- Examp-{J As tapproaches 0 from the left, H[t) approaches 0 . As I approaches 0 from the right, H(t} approaches El . Therefore the limit as t approaches 0 of H{t) does not exist. Nee-mew? 6. -;'1 paintsSEECElcEl'Z 1.3.035. Prove the statement using the 8, 6 denition ofa limit. um M2? x13 x3 Given 8 3- 0, we need 6 IF;Q_ a such that if 0 < Ix 3| : 5, then x2 +x _ 12 7 x3 De <86: |x+47| x2+x12_7 x3 Need Help? 7. 1'1 paintsSEECIllcm 1.3.503.XP. Explain the meaning of each of the following. [a] lim x) = an x 1 8 The values of fix) can be made arbitrarily large by taking x sufciently close to (but not equal to) 8. (:1 The values of fix) can be made arbitrarily close to 8 by taking x sufciently large. cw 118) = on O The values of x} can be made arbitrarily close to O by taking x sufciently close to {but not equal to) 8. (b) lim f(x} = m x - 5+ O The values of fix) can be made arbitrarily close to c0 by taking it sufciently close to 5. Me values of x) can be made negative with arbitrarily large absolute values by taking x sufciently close to, but greater than, 5. C} {(5} z oo C) As x approaches 5, for) approaches m. Need Help? ii 8. -;'l. pointSSEECalclzI'Z 1.3.531.XP. Prove the statement using the 3L 6 denition of a limit. I' _ = X _I_m_3 (x? 1) 3 Given 8 3 0, we need 3 r50 ' a such that if 0 4: Ix (3\" < 3, then |[)(2 1} 8| leg 3 or upon simplifying weneed |x29|<8 whenever 0< |x+3| <5. Noticethatif |x+3|<1,then1 7|x3| 0, we need 13 + 3x 4 < 4 = such that if 0 < |x 1| < , then >0 3x 3 4 (4/3)! < 3 |x 1| < 4 , then 0 < |x 1| < |x 1| < 13 + 3x 4 13 + 3x 4 4 . So if we choose (4/3)! 4 < . Thus, . But 0, we need (x + 4)(x 3) 7 < x3 Then 0 < |x 3| < such that if 0 < |x 3| < , then >0 |x + 4 7| < [x 3] 0 < |x 3| < x2 + x 12 7 x3 0 < |x 3| < . So choose = 0 < |x + 4 7| < ! . (x + 4)(x 3) 7 < [x 3] x3 x2 + x 12 7 < . By the definition of a limit, lim x2 + x 12 = 7 x3 x3 x3 https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 0, we need such that if 0 < |x (3)| < , then |(x2 1) 8| >0 or upon 0, we need > 0 such that if 0 < |x 0| < , then |x2 0| < . < = |x| < 0 Take . Then 0 < |x 0| < |x2 0| < . Thus, lim x2 = 0 by x0 the definition of a limit. https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1. -/1 pointsSEssGlcFl'Z 1.3.001. Ifa ball is thrown in the air with a velocity 46 Ft/s, its height in feet 1' seconds later is given by y = 461' 1612. (a) Find the average velocityIr for the time period beginning when t = 2 and lasting 6) 0.5 second. Ems Oi) 0.1 second. Elm ii) 0.05 second. ws UV) 0.01 second. ftls (in) Estimate the instantaneous velocity when t = 2. [El ws Need Help? i i i \f\fW m m m m follqwlng : llrn 9M x-I'Z, : xljgx) (b) 312,me (c) (a) mm M 5. (-1) some" From the graph we see maybe yams (c) Slncg che le: and \"gm: nm 9x151; 1.3 Homework ! : // WebAssign jarallah almarri MAT 265, section 72204, Fall 2017 Instructor: Sergey Nikitin 1.3 Homework (Homework) Current Score : 10.28 / 15 Due : Saturday, August 26 2017 11:59 PM MST 1. -/1 pointsSEssCalcET2 1.3.001. If a ball is thrown in the air with a velocity 46 ft/s, its height in feet t seconds later is given by y = 46t 16t2. (a) Find the average velocity for the time period beginning when t = 2 and lasting (i) 0.5 second. ft/s (ii) 0.1 second. ft/s (iii) 0.05 second. ft/s (iv) 0.01 second. ft/s (b) Estimate the instantaneous velocity when t = 2. ft/s https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1.3 Homework ! : // 2. 0.76/1 points | Previous AnswersSEssCalcET2 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) lim g(t) lim g(t) t 0 -1 (b) t 0+ -2 (c) lim g(t) t0 0 (d) lim g(t) lim g(t) t 2 2 (e) t 2+ 0 (f) lim g(t) t2 -2 (g) g(2) 1 (h) lim g(t) t4 3 Enhanced Feedback Please try again. For finding limits, the function value at t = a is irrelevant. Simply approach the point t = a from the left (if lim t a is requested), or from the right (if lim t a+ is requested). For lim t a, take left and right limits and make sure that they are the same. https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1.3 Homework ! : // 3. 1/1 points | Previous AnswersSEssCalcET2 1.3.007. Sketch the graph of an example of a function f that satisfies all of the given conditions. lim x 0 f(x) = 1, lim x 0+ f(x) = 2, f(0) = 1 https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1.3 Homework ! : // 4. 0/1 points | Previous AnswersSEssCalcET2 1.3.023. x to find a number such that Use the given graph of f(x) = |x 4| < if = then x 2 < 0.4. 1.76 Enhanced Feedback Please try again. For finding such that |x a2| < implies | x a| < b, start by finding the solutions to | Choose so that neither of these solutions are at a distance smaller than x a| = b. from a2. 5. 1/1 points | Previous AnswersSEssCalcET2 1.3.033. Prove the statement using the , definition of a limit. lim x1 13 + 3x =4 4 Given > 0, we need 13 + 3x 4 < 4 = such that if 0 < |x 1| < , then >0 3x 3 4 (4/3)! < 3 |x 1| < 4 , then 0 < |x 1| < |x 1| < 13 + 3x 4 13 + 3x 4 4 . So if we choose (4/3)! 4 < . Thus, . But 0, we need (x + 4)(x 3) 7 < x3 Then 0 < |x 3| < such that if 0 < |x 3| < , then >0 |x + 4 7| < [x 3] 0 < |x 3| < x2 + x 12 7 x3 0 < |x 3| < . So choose = 0 < |x + 4 7| < ! . (x + 4)(x 3) 7 < [x 3] x3 x2 + x 12 7 < . By the definition of a limit, lim x2 + x 12 = 7 x3 x3 x3 https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 0, we need such that if 0 < |x (3)| < , then |(x2 1) 8| >0 or upon 0, we need > 0 such that if 0 < |x 0| < , then |x2 0| < . < = |x| < 0 Take . Then 0 < |x 0| < |x2 0| < . Thus, lim x2 = 0 by x0 the definition of a limit. https://www.webassign.net/web/Student/Assignment-Responses/last?dep=16578673 1. -/1 pointsSEssGlcFl'Z 1.3.001. Ifa ball is thrown in the air with a velocity 46 Ft/s, its height in feet 1' seconds later is given by y = 461' 1612. (a) Find the average velocityIr for the time period beginning when t = 2 and lasting 6) 0.5 second. Ems Oi) 0.1 second. Elm ii) 0.05 second. ws UV) 0.01 second. ftls (in) Estimate the instantaneous velocity when t = 2. [El ws Need Help? i i i \f\fW m m m m follqwlng : llrn 9M x-I'Z, : xljgx) (b) 312,me (c) (a) mm M 5. (-1) some" From the graph we see maybe yams (c) Slncg che le: and \"gm: nm 9x151