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Unformatted text preview: MAT 131 FINAL EXAM – SOLUTIONS FALL 2010 (1) (a) The limit is the indeterminate form 0 / 0. Using L’Hospital’s Rule: lim x → sin(2 x ) sin(3 x ) = lim x → 2cos(2 x ) 3cos(3 x ) = 2 · 1 3 · 1 = 2 3 (b) The limit is the indeterminate form 0 / 0. Using L’Hospital’s Rule and the known limit lim x → sin( x ) /x = 1: lim x → 1 cos( x ) x 2 = lim x → sin( x ) 2 x = 1 2 lim x → sin x x = 1 2 (c) Using the factorizations for a 3 b 3 and a 2 b 2 : lim x → 1 x 3 x 3 x 2 x 2 = lim x → 1 ( x x 1 )( x 2 + xx 1 + x 2 ) ( x x 1 )( x + x 1 ) = lim x → 1 x 2 + 1 + x 2 x + x 1 = 3 2 (d) First we find the limit of ln x 1 / ( x 1) = (ln x ) / ( x 1), using L’Hospital’s Rule: lim x → 1 ln x x 1 = lim x → 1 1 /x 1 = 1 . Then, because x 1 / ( x 1) = e (ln x ) / ( x 1) and the exponential function is continuous, lim x → 1 x 1 / ( x 1) = lim x → 1 e (ln x ) / ( x 1) = e 1 = e (2) y = f ( x ), with y = 6 x 2 8 x 3 = 6 x 1 8 x 3 . (a) Computation of f ( x ) and f 00 ( x ), using the power rule and the sum rule: f ( x ) = 6 x 2 + 24 x 4 f 00 ( x ) = 12 x 3 96 x 5 (b) f ( x ) is odd , because f ( x ) = 6( x ) 2 8 ( x ) 3 = 6 x 2 8 x 3 = 6 x 2 8 x 3 = f ( x ) (c) Vertical asymptotes occur where the denominator of the rational expression defining y equals zero. The equation of the only vertical asymptote is x = 0 . (d) Horizontal asymptotes are found by looking at the limit of the function at ±∞ . lim x →∞ 6 x 2 8 x 3 = lim x →∞ 6 x 2 8 x 3 = 0 because the degree of the denominator is greater than the degree of the numer ator. The only horizontal asymptote isator....
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This note was uploaded on 01/13/2012 for the course MAT 131 taught by Professor Christopherbay during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 CHRISTOPHERBAY
 Calculus

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