Limits, Continuity, and Differentiability Solutions We have intentionally included more material than can be covered in most Student Study Sessions to account for groups that are able to answer the questions at a faster rate. Use your own judgment, based on the group of students, to determine the order and selection of questions to work in the session. Be sure to include a variety of types of questions (multiple choice, free response, calculator, and non-calculator) in the time allotted. Quick Check for Understanding: Student graphs will vary. Sample answers given.. Sketch a function with the property that f (a) eists but lim f () does not eist. y a 0 0 0 0 0 0. Sketch a function with the property that lim f () eists but f (a) does not eist. a 0 0 0 0 y 3. Sketch a function with the property that f (a) eists and lim f () eists but lim f () f (a). a a 0 y 0 0 0 Copyright 03 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at www.nms.org 0 0 y
Multiple Choice Limits, Continuity and Differentiability. D (95 AB5) Since the limit is taken as n and the eponents in the numerator and denominator are n equal, use the ratio of the leading coefficients to find that the limit is n.. B (00 AB) Multiplying in the numerator and denominator yields the equivalent limit: 73 lim. 3 3. B (99 AB/BC) This limit represents the definition of the derivative of the function, f ( ) = 7 ; f =. f ( ) at.. D (95 BC9 appropriate for AB) sin sin Let t. lim t lim. t0 t 5. B (9 AB9) This limit represents the derivative of the function, f ( ) tan3. Using the chain rule, f ( ) 3sec (3 ). B (993 BC appropriate for AB) lim 0 7. B (9 AB3) Since f ( ) cos, f ( ) sin. Also f (0) 0, g(0) 0, and g( ), hence f( ) sin lim lim. 0 g ( ) 0 f( ) f( ) f(0) cos0 Alternatively, by L Hopital s rule, lim lim. 0 g ( ) 0 g( ) g(0). C (993 AB9) cos cos cos lim lim lim lim. 0 0 sin ( cos 0 ( cos )( cos ) 0 ( cos ) cos sin Alternatively, by L Hopital s rule, lim lim lim. 0 sin 0 sin cos 0 cos. Copyright 03 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at www.nms.org
Limits, Continuity, and Differentiability 9. E (95 AB) Using the given limit, there is not enough information to establish that f ( a) eists, nor that f ( ) is continuous or defined at a, nor that f ( a) L. For eample, consider the function whose graph is the horizontal line y with a hole at a. For this function lim f( ) and none of the given statements are true a 0. E (003 AB3) By definition of a horizontal asymptote, E is correct.. E (993 AB35) Since y is a horizontal asymptote, the ratio of the leading coefficients must be ; therefore, a. Since there is a vertical asymptote at 3, set the denominator, 3 c, equal to 0, so c3then ac3 5.. E (9 AB7) f (3) (3) 9 9 lim lim 9 9 3 3 Since f (3) lim f( ) the function is continuous at 3 3, 3 f( ), 3 and lim lim 3 3 Since the left and right limits of the derivative of the function are equivalent from either side of 3, the function is differentiable at 3. 3. D (003 AB0) Using substitution, the one sided limits as 3 are both equal to 5; therefore, I and II are true. f is not differentiable at 3 since f (3) for 3 and f (3) for 3.. A (00 AB/BC) Use the top piece of the piecewise function for the limit since the bottom piece gives the ( )( ) value of f (). Using factoring, lim lim lim. Since this limit does not equal f () =, the function f is not continuous or differentiable at. 5. C (9 BC5 appropriate for AB) lim f ( ) f ( a ) for all values of a ecept. a lim f ( ) lim( ) 0 f (). Copyright 03 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at www.nms.org
Limits, Continuity, and Differentiability. E (99 AB/BC) Thinking graphically, the absolute value function will have a sharp corner at 3, thus the derivative does not eist at that point. The question can also be worked algebraically as follows: The one sided limits are not equal: f( ) f(3) 3 lim 3 3 3 f( ) f(3) 3 lim. 3 3 3 Therefore, the value of f (3) does not eist. 7. A (993 AB5) Consider f ( a) lim f( ) a Use factoring to simplify the function and then substitute for : ( )( ) lim lim ; therefore, f ( ).. B (997 AB5) The left and right limits as a are both equal to. The limit as b does not eist since the one-sided limits are not equivalent; therefore, A, C and D cannot be true either. 9. A (003 AB3/BC3) The graph of f is continuous at a ; however, since the graph has a sharp turn at a, the function is not differentiable at a. 0. D (003 AB79) The one-sided limits as are equivalent for the graphs of f in I and II but not for III.. C (00 AB77) lim f ( ) and lim f ( ) eist. eist; however, since they are not equivalent, the lim f ( ) does not. A (003 BC appropriate for AB) sin() 0.9093 Copyright 03 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at www.nms.org
Free Response 3. 009B AB3abc/BC3abc (a) f( h) f(0) lim 0 h 3 f( h) f(0) lim 0 0 h Since the one-sided limits do not agree f is not differentiable at 0. f() f( a) (b) 0 when f ( a) f(). There a are two values of a for which this is true. (c) Yes, a 3. The function f is differentiable on the interval 3 and continous on f() f(3) 0 3. Also,. 3 3 3 By the Mean Value Theorem, there is a value c, 3, such that f() c. 3 Limits, Continuity, and Differentiability : sets up difference quotient at 0 : answer with justification : epression for average rate of change : answer with reason : answers yes and identifies a 3 : justification. 003 ABac (a) f is continuous at 3 because lim f lim f. 3 3 Therefore, f f lim 3. 3 : answer yes and equates the values of the left- and righthand limits : eplanation involving limits (c) Since g is continuous at 3, k 3m. k ; 0 3 g m ; 3 5 k lim g and 3 lim g m 3 Since these two limits eist and g is differentiable at 3, the two limits are k equal. Thus m. m3m; m and k 5 5 : k 3m k 3 : m : values for k and m Copyright 03 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at www.nms.org
5. 0 ABab lim sin (a) 0 lim e 0 f 0 So, lim f f 0 0. Therefore f is continuous at 0. Limits, Continuity, and Differentiability : analysis cos for 0 e for 0 cos 3 for all values of 0. 3 e 3 when ln 0. 3 Therefore f 3 for ln. (b) f 3 : f : value of Student samples for question 5a Sample A: Student earned both points for part (a). The student clearly identified the three attributes required to be true to justify continuity. Sample B: Student earned point for part (a). The student begins the analysis of continuity by looking at the functional values on each side of. The student does not use limits and does not consider f (0), thus earning only of the possible points. Copyright 03 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at www.nms.org