MATH 203 Lecture Notes - Lecture 3: Farad, Asymptote

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10 May 2016

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Class 3_203: limits at in(cid:133)nity and horizontal asymptotes: (a) we (should) know here that: p > 0 : lim x! (cid:6)1 (b) f (x) = 1 xp = 0, if p = 0 : 3x + 5 px2 + 1 lim x!1 lim x! (cid:6)1. 1 xp = 1; p < 0 : (3x + 5) (cid:4) x px2 + 1 (cid:4) x. 3x + 5 px2 + 1 px2 + 1 since x lim x! (cid:6)1. 3 therefore y = 3 is a horizontal asymptote. lim x! (cid:0)1 lim x! (cid:0)1. = lim x! (cid:0)1 since (3x + 5) (cid:4) x px2 + 1 (cid:4) x(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) xr1 + = (cid:0)3 px2 + 1 x px2 + 1 px2 (cid:0) = (cid:0)r x2 + 1 x2 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y. 1 (cid:4) x px2 + 1 (cid:4) x. 0 therefore y = 0 is a horizontal asymptote.

Related Questions

[8] 9. Either prove the statement is true or show that it is false. (a) sin (sin(2)) = 27. Solution: It is FALSE. The range of sin-' r is (-1,1], hence sin (sin(27)) = sin (0) = 0 (b) If f is continuous at a, then f is continuous at a. Solution: It is TRUE. Let g(x) = \f()]. If lim f(x) = f(a), then we have since the absolute value function is continuous lim g(x) = lim f() = lim f(x) = f(a) = g(a) So, g = \f is continuous at a. (c) If [g] is continuous at a, then g is continuous at a. -1 if r <0 Solution: It is FALSE. Let g(x) = 3 · Then, g = 1, so g is continuous at 0 i if >0 by the Continuity Theorems, but lim g(t) = -1 + lim g(1) 240- 1- 0+ So, g(x) is not continuous at a. (d) If f is continuous at a, then f is differentiable at a. Solution: It is FALSE. We know that f(1) = 2 is continuous at 0, but is not differentiable at 0.

4. (16 pts) MULTIPLE CHOICE! CIRCLE THE CORRECT ANSWER IN EACH PART. (I) ( 4 pts) Complete the statement of the Intermediate Value Theorem: Suppose f is continuous on the interval [a, b] and L is a number between f (a) and f (b). Then there is at least one number c in (a, b) such that (a) C = L; (b) f(c) = 0; (c) f(c) = L; (a) f' (c) = L; (e) C = 0; (f) f (a) < c < f (b); (g) NONE OF THE PREVIOUS ANSWERS. (II) (4 pts) The figure shows the graph of a function f. At what point (points) c does the conclusion of the Intermediate Value Theorem hold for f (x) on the interval [1, 5] and L = 2. --- - - - -1 - - - 2 - - I- - - (a) c = 1; (b) C = 0; (c) C = 2; (a) c = 3; (e) C = 4; (f) C = 5; (g) c= -1. (III) (4 pts) Given that cos x s f (x) s et, for all x > 0 evaluate lim f (x) x+0+ (a) lim f (x) = 0; (b) lim f (x) = 1; (c) lim f (x) = e; x0+ X0+ (d) lim f(x) DOES NOT EXIST. WE DON'T HAVE ENOUGH INFORMATION TO ANSWER THE QUESTION *40+ (IV) (4 pts) Given the function 1 f (x) = , find its inverse, f+(x) x -2 (a) f (x) = x -2; (b) f«+ (x) = x+2 ; (c) fº+ (x) = -2; (f) f-+ (x) -2; (e) f (x) DOES NOT EXIST.