Final Exam from Last Year

# Final Exam from Last Year - MAT 131 FINAL EXAM SOLUTIONS...

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MAT 131 FINAL EXAM – SOLUTIONS FALL 2010 (1) (a) The limit is the indeterminate form 0 / 0. Using L’Hospital’s Rule: lim x 0 sin(2 x ) sin(3 x ) = lim x 0 2 cos(2 x ) 3 cos(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 0 sin( x ) /x = 1: lim x 0 1 - cos( x ) x 2 = lim x 0 sin( x ) 2 x = 1 2 lim x 0 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 0 ( x ) and f 00 ( x ), using the power rule and the sum rule: f 0 ( 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 is y = 0 .

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