MIT18_014F10_ex3_sols

MIT18_014F10_ex3_sols - Exam 3 Solutions Problem 1....

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Unformatted text preview: Exam 3 Solutions Problem 1. Evaluate lim x x 1 1 . log( x +1) Solution We begin by writing the problem as a single fraction, lim x log( x +1) x . x log( x +1) Observe that both numerator and denominator have limit zero, and thus we can apply LHopitals rule to see lim log( x + 1) x = lim 1 / ( x + 1) 1 = lim 1 ( x + 1) . x x log( x + 1) x log( x + 1) + x/ ( x + 1) x ( x + 1) log( x + 1) + x Note that in the expression on the right, the limits of both numerator and denominator are again zero. Thus a second application of LHopitals rule gives 1 ( x + 1) 1 1 x lim ( x + 1) log( x + 1) + x = lim x log( x + 1) + ( x + 1) / ( x + 1) + 1 = 2 . 3 x 2 Problem 2. Evaluate x 2 6 x +10 dx. Solution We start by observing that the denominator can be written as ( x 3) 2 + 1. That makes part of the problem easy: dx 2 ( x 3) 2 + 1 = 2 arctan( x 3) ....
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MIT18_014F10_ex3_sols - Exam 3 Solutions Problem 1....

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