Cos x ² lim x sin x x ² lim x 1 cos x 1 ² lim x

This preview shows page 2 - 5 out of 12 pages.

cos x ² lim x ! 0 sin x x ² lim x ! 0 1 cos x ) 1 ² lim x ! 0 sin x x ² 1 (i.e. lim x ! 0 g ( x ) = 1 ) (d) Use the result in part (c) to determine lim x ! 0 sin x sin (3 x ) : lim x ! 0 sin x sin (3 x ) = lim x ! 0 ²³ sin x x ± x ´ ± ³ 3 x sin(3 x ) ± 1 3 x ´µ = 1 3 ± lim x ! 0 sin x x ± lim x ! 0 3 x sin(3 x ) = 1 3 ± 1 ± 1 = 1 3 Grade: 24 24
MA103 Lab Report 2 - Limit Applications Would you like this assignment Graded? YES or NO Name: Student Number: Fall 2016 1. [4 marks ] Recall Question #2, Lab Report 1, where f ( x ) = p x 2 ° 9 ° 4 x ° a and a was constant. Determine all horizontal asymptotes of f ( x ) : lim x !1 f ( x ) = lim x !1 " p x 2 ° 9 ° 4 x ° a ± p x 2 ± x !# lim x !°1 f ( x ) = lim x !°1 " p x 2 ° 9 ° 4 x ° a ± ° p x 2 ± x !# = lim x !1 " p 1 ° 9 =x 2 ° 4 = p x 2 1 ° a=x # = lim x !°1 " ° p 1 ° 9 =x 2 + 4 = p x 2 1 ° a=x # = p 1 ° 0 ° 0 1 ° 0 = 1 = ° p 1 ° 0 + 0 1 ° 0 = ° 1 ) f has HAs at y = ° 1 and y = 1 : 2. [3 marks ] Suppose that Canada°s population (in millions) can be modelled by L ( t ) = 120 e t= 50 e t= 50 + 21 , where t = 0 is the year 1900.Determine thelim 120 e t= 50 e t= 50 + 21 ° ± e t= 50 ± e t= 50 ± 120 1 + 21 =e t= 50 3. [11 marks ] Consider the function f ( x ) = 8 < : p x 2 + 16 ; x ² ° 3 4 x ° k x ; x > ° 3 where k is constant. (a) De±ne f in Maple and evaluate each of the following limits. f:=(x)- > piecewise(x < =-3,sqrt(x^2+16),(4*x-k)/x); limit(f(x),x=-3,right); lim x 3 + f ( x ) = 4 + 1 3 k limit(f(x),x=-3,left); lim x 3 ° f ( x ) = 5 (b) For what value of k would f ( x ) be continuous at x = ° 3? To be continuous at x = ° 3 : lim x 3 + f ( x ) = lim x 3 ° f ( x ) = f ( ° 3) = 5 ) 4 + 1 3 k = 5 ) k = 3 (5 ° 4) = 3 (c) Using the result for k in part (b), plot f ( x ) for x 2 [ ° 10 ; 10] and y 2 [ ° 10 ; 10] in Maple : k:=???; plot(f(x),x=-10..10,y=-10..10,discont=true); (substitute appropriately for ???) Analyzing the graph, state whether f ( x ) x ° 3 0 3 is continuous and/or di/erentiable at Continuous? (Y or N) Y N Y each of the following points : Di/erentiable? (Y or N) N N Y
(d) Use the limit de±nition of a derivative to determine the value of the derivative at x = 2 : [Note: You may assume k is the value determined in part (b).] f 0 (2) = lim h ! 0 f (2 + h ) ° f (2) h = lim h ! 0 1 h ² 4(2 + h ) ° 3 (2 + h ) ° 4(2) ° 3 2 ³ = lim h ! 0 1 h ² 4 h + 5 h + 2 ° 5 2 ³ = lim h ! 0 1 h ² 2(4 h + 5) ° 5( h + 2) 2( h + 2) ³ = lim h ! 0 1 h ² 3 h 2 h + 4 ³ = lim h ! 0 3 2 h + 4 = 3 4 (e) Explain whether the Intermediate Value Theorem could be used to verify that f has a root (i.e. f ( c ) = 0 ) on each of the following intervals. [Hint: Ensure the conditions of IVT are satis±ed.] i. [ ° 1 ; 3] f is discontinuous x = 0 ; so IVT cannot be used to show there is a root on [ ° 1 ; 3] . ii. ² 1 2 ; 3 2 ³ f is continuous on ² 1 2 ; 3 2 ³ and f ° 1 2 ± = ° 2 < 0 < f ° 3 2 ± = 2 ; so IVT concludes that there is a root on ² 1 2 ; 3 2 ³ . 4. [3 marks ] Suppose that the sales revenue for a new smart phone depends on its selling price p ³ 0 (in dollars) and is given by R ( p ) = 8200 p 250000 + p 2 (in billions of dollars).

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture