021midtermsoln2

021midtermsoln2 - HKUST MATH021 Concise Calculus Midterm...

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HKUST MATH021 Concise Calculus Midterm Solution (Yellow Version) 1. Find limits (a) lim x 0 p 3 x + 1 + x x 2 + 1 - 1 x . [4 points] Solution lim x 0 p 3 x + 1 + x x 2 + 1 - 1 x = lim x 0 ( p 3 x + 1 + x x 2 + 1 - 1)( p 3 x + 1 + x x 2 + 1 + 1) x ( p 3 x + 1 + x x 2 + 1 + 1) = lim x 0 (3 x + 1 + x x 2 + 1) - 1 x ( p 3 x + 1 + x x 2 + 1 + 1) = lim x 0 3 x + x x 2 + 1 x ( p 3 x + 1 + x x 2 + 1 + 1) = lim x 0 3 + x 2 + 1 p 3 x + 1 + x x 2 + 1 + 1 = 3 + 1 1 + 1 = 2 (b) lim x 1 ln h 1 + sin ± π 2 p x 2 - x + 1 ²i . [4 points] Solution lim x 1 ln h 1 + sin ± π 2 p x 2 - x + 1 ²i = ln h 1 + sin ± π 2 p 1 2 - 1 + 1 ²i = ln(1 + sin π 2 ) = ln(1 + 1) = ln 2 (c) lim x 1 - e 1 1 - x . [4 points] Solution lim x 1 - e 1 1 - x = e = (d) lim x 1 ( x - 1) 2 sin 1 x 2 - 1 . [4 points] Solution From - 1 sin 1 x 2 - 1 1 , we have - ( x - 1) 2 ( x - 1) 2 sin 1 x 2 - 1 ( x - 1) 2 . Since lim x 1 [ - ( x - 1) 2 ] = lim x 1 ( x - 1) 2 = 0 , using the Squeeze Theorem (the Sandwich Theorem), we have lim x 1 ( x - 1) 2 sin 1 x 2 - 1 = 0 .
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1 2. Find the derivatives of the following functions. (a) y = sin x x + sin x [4 points] Solution dy dx = ( x + sin x ) d sin x dx - sin x d ( x + sin x ) dx ( x + sin x ) 2 = ( x + sin x ) cos x - sin x (1 + cos x ) ( x + sin x ) 2 = x cos x - sin x ( x + sin x ) 2 (b) y = ( x 2 + x + 3) e - 3 x [4 points] Solution dy dx = ( x 2 + x + 3) de - 3 x dx + e - 3 x d ( x 2 + x + 3) x = ( x 2 + x + 3)( - 3 e - 3 x ) + e - 3 x (2 x + 1)
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This note was uploaded on 02/27/2010 for the course MATH MATH021 taught by Professor Nil during the Fall '08 term at HKUST.

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021midtermsoln2 - HKUST MATH021 Concise Calculus Midterm...

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