prelin spring 2005 solutions

prelin spring 2005 solutions - Prelim 1 Solutions timothy...

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Prelim 1 Solutions timothy goldberg Spring 2005 (1) (a) This is not a Squeeze Theorem problem. The function sin x x is continuous everywhere it is defined, which includes the number π / 2. Therefore lim x π / 2 sin x x = sin( π / 2) π / 2 = 1 π / 2 = 2 / π . (b) We solve this problem by factoring: lim x 1 x 2 - 1 xe x - e x = lim x 1 ( x - 1)( x + 1) e x ( x - 1) = lim x 1 x + 1 e x = 2 /e. (c) This is a Squeeze Theorem problem. We know - 1 cos x 1, so - 1 x + 1 cos x x + 1 1 x + 1 . Because lim x →∞ - 1 x +1 = 0 and lim x →∞ 1 x +1 = 0, the Squeeze Theorem implies that lim x →∞ cos x x + 1 = 0 . (There are other ways to write down a solution to this problem, but this is the basic idea.) (2) (a) There are two di ff erent limit definitions for the derivative. Using one of them, we have f (3) = lim x 3 f ( x ) - f (3) x - 3 = lim x 3 x + 1 - 3 + 1 x - 3 = lim x 3 x + 1 - 2 x - 3 = lim x 3 x + 1 - 2 x - 3 · x + 1 + 2 x + 1 + 2 = lim x 3 x + 1 2 - 2 x + 1 + 2 x + 1 - 4 ( x - 3)( x + 1 + 2) = lim x 3 x + 1 - 4 ( x - 3)( x + 1 + 2) = lim x 3 x - 3 ( x - 3)( x + 1 + 2) = lim x 3 1 x + 1 + 2 = 1 3 + 1 + 2 = 1 4 . 1
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Using the other limit definition, we have f (3) = lim h 0 f (3 + h ) - f (3) h = lim h 0 3 + h + 1 - 3 + 1 h = lim h 0 h + 4 - 2 h = lim h 0 h + 4 - 2 h · h + 4 + 2 h + 4 + 2 = lim h 0 h + 4 2 - 2 h + 4 + 2 h + 4 - 4 h ( h + 4 + 2) = lim h 0 h + 4 - 4 h ( h + 4 + 2) = lim h 0 h h ( h + 4 + 2) = lim h 0 1 h + 4 + 2 = 1 0 + 4 + 2 = 1 4 . (b) Note that this equation is the same as the function f ( x ) from part (a). The tangent line to f ( x ) at x = 3 has equation y - f (3) = f (3)( x - 3) . From part (a) we know f (3) = 1 / 4, and f (3) = 3 + 1 = 2, so this tangent line equation becomes y - 2 = 1 4 ( x - 3) . (3) (a) Noting that e π is a constant and hence has derivative equal to 0, we calculate d dx [ e x + x π + e π ] = d dx [ e x ] + d dx [ x π ] + d dx [ e π ] = e x + π x π - 1 .
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