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Assignment 5 Solutions.pdf

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Department
Calculus
Course
Calculus 1000A/B
Professor
Chris Brandl
Semester
Fall

Description
MATH 127 Fall 2012 Assignment 5 Solutions 1. Evaluate each of the following limits 11x (a) lim p x!▯1 x + 4x 11x 11x lim p = lim q ▯ ▯ x!▯1 x + 4x x!▯1 x2 1 +4x x2 11x = lim q x!▯1 p 2 4 x 1 + x 11x = x!▯1 q jxj 1 + x 11x = lim q x!▯1 ▯x 1 + 4 x ▯11 = lim q x!▯1 4 1 +x r ! 4 = ▯11 note that lim 1 + = 1 x!▯1 x 3e + e▯x (b) lim x!▯1 e ▯ e▯x+1 3e + e▯x 3e + x lim = lim e x!▯1 e ▯ e▯x+1 x!▯1 e ▯ ex 2x 3ex+1 = lim 2x x!▯1 ex▯e e 3e2x+ 1 = lim 2x x!▯1 ▯ ▯ e ▯ 1 = ▯ since lim 3e2x= 0 and lim e2x= 0 e x!▯1 x!▯1 ▯x p (c) lim e sin x + 2 x!1 p p Note that ▯1 ▯ sin x + 2 ▯ 1 so we get that ▯ex ▯ e▯xsin x + 2 ▯ e▯x. ▯x ▯x p Since lim ▯e = 0 we get by the Squeeze Theorem that lim sin x + 2 = 0. x!1 x!1 1 p 2 (d) lim 4x ▯ 2x + 1 ▯ 2x x!1 ▯ p ▯ p ▯p ▯ 4x ▯ 2x + 1 + 2x lim 4x ▯ 2x + 1 ▯ 2x = lim 4x ▯ 2x + 1 ▯ 2x p x!1 x!1 4x ▯ 2x + 1 + 2x 2 2 = lim px ▯ 2x + 1 ▯ 4x x!1 4x ▯ 2x + 1 + 2x ▯2x + 1 = x!1 p 2 4x ▯ 2x + 1 + 2x ▯2x + 1 = lim q ▯ ▯ x!1 x2 4 ▯ 2+ 1 + 2x x x2 ▯2x + 1 = lim q x!1 2 1 jxj 4 ▯ x+ x2+ 2x = lim q ▯2x + 1 x!1 2 1 x 4 ▯x + x2+ 2x ▯ ▯ x ▯2 + x = lim ▯q ▯ x!1 x 4 ▯ 2+ 1 + 2 x x 1 ▯2 + x = x!1 q 2 1 4 ▯x + x2+ 2 ▯2 = p 4 + 2 1 = ▯ 2 2. For the following functions, using the de▯nition of the derivative either ▯nd the indi- cated derivative or show the derivative does not exist. 8 > < x x < 0 (a) f (0) where f(x) = : sin(x) x ▯ 0 0 f(x) ▯ f(0) We need to compute f (0) = lim but since f is de▯ned di▯erently on x!0 x ▯ 0 2 either side of x = 0 we must observe left and right hand limits. f(x) ▯ f(0) x ▯ sin(0) li▯ = lim▯ x!0 x x!0 x x = x!0▯x = 1 f(x) ▯ f(0) sin(x) ▯ sin(0) li+ = lim+ x!0 x x!0 x sin(x) = lim+ x!0 x = 1 f(x) ▯ f(0) f(x) ▯ f(0) Since lim = 1 = lim we get that x!0+ x x!0▯ x 0 f(x) ▯ f(0) f (0) = lim = 1 x!0 x 0 1 (b) f (x) where f(x) 1 + x f (x) = limx + h) ▯ f(x) h!0 h 1 1 p1+x+h▯ p1+x = lim h!0p h p 1 + x ▯ 1 + x + h = lim p p h!0h( 1 + x + h 1 + x) ▯ p p ▯▯ p p ▯ 1 + x ▯ 1 + x + h 1 + x + 1 + x + h = h!0 p p p p h( 1 + x + h 1 + x) 1 + x + 1 + x + h 1 + x ▯ (1 + x + h) = lim p p h!0h 1 + x + h(1 + x) + h(1 + x + h) 1 + x = lim p ▯h p h!0h 1 + x + h(1 + x) + h(1 + x + h) 1 + x ▯1 = h!0p p 1 + x + h(1 + x) + (1 + x + h) 1 + x ▯1 = p p 1 + x(1 + x) + (1 + x) 1 + x ▯1 = 3 2(1 + x) 3 (c) f (x) where f(x) = x ▯ 3x 0 f(x + h) ▯ f(x) f (x) = lim h!0 h
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