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021midtermsoln1

021midtermsoln1 - HKUST MATH021 Concise Calculus Midterm...

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HKUST MATH021 Concise Calculus Midterm Solution (White Version) 1. Find limits (a) lim x 0 p x + 1 + 2 x x 2 + 1 - 1 x . [4 points] Solution lim x 0 p x + 1 + 2 x x 2 + 1 - 1 x = lim x 0 ( p x + 1 + 2 x x 2 + 1 - 1)( p x + 1 + 2 x x 2 + 1 + 1) x ( p x + 1 + 2 x x 2 + 1 + 1) = lim x 0 ( x + 1 + 2 x x 2 + 1) - 1 x ( p x + 1 + 2 x x 2 + 1 + 1) = lim x 0 x + 2 x x 2 + 1 x ( p x + 1 + 2 x x 2 + 1 + 1) = lim x 0 1 + 2 x 2 + 1 p x + 1 + 2 x x 2 + 1 + 1 = 1 + 2 1 + 1 = 3 2 (b) lim x 1 ln h 1 + cos ± π 2 p x 2 + 2 x - 2 ²i . [4 points] Solution lim x 1 ln h 1 + cos ± π 2 p x 2 + 2 x - 2 ²i = ln h 1 + cos ± π 2 p 1 2 + 2 · 1 - 2 ²i = ln(1 + cos π 2 ) = ln(1 + 0) = 0 (c) lim x 1 + e 1 1 - x . [4 points] Solution lim x 1 + e 1 1 - x = e -∞ = 0 (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 = cos x x + cos x [4 points] Solution dy dx = ( x + cos x ) d cos x dx - cos x d ( x + cos x ) dx ( x + cos x ) 2 = - ( x + cos x ) sin x - cos x (1 - sin x ) ( x + cos x ) 2 = - x sin x - cos x ( x + cos x ) 2 (b) y = ( x 2 + 2 x + 3) e - 2 x [4 points] Solution dy dx = ( x 2 + 2 x + 3) de - 2 x dx + e - 2 x d ( x 2 + 2 x + 3) x = ( x 2 + 2 x + 3)( - 2 e - 2 x ) + e - 2 x (2 x + 2) = - 2( x 2 + x + 2) e - 2 x (c) y = 1 7 x 2 + 3 x + 2 [4 points] Solution dy dx = d dx ( x 2 + 3 x + 2) - 1 7 = - 1 7 ( x 2 + 3 x + 2) - 1 7 - 1 d ( x 2 + 3 x + 2) dx = - 2 x + 3 7 ( x 2 + 3 x + 2) - 8 7 (d) y = ( x + 3) x 2 +4 [4 points] Solution
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021midtermsoln1 - HKUST MATH021 Concise Calculus Midterm...

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