exam2_2009_sol

| s(3 s(1 | = | 2 | | 18 2 | = 22 meters 3 evaluate

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Unformatted text preview: | s (3)- s (1) | = |- 2- | + | 18- (- 2) | = 22 meters. 3. Evaluate the following limits, or explain why they do not exist. (a) [6 points] lim x ! 2 x- 2 p 5 + x 2- 3 . Math 171: Exam 2 Solution: Multiplying and dividing by the conjugate: lim x ! 2 x- 2 p 5 + x 2- 3 = lim x ! 2 ( x- 2) ( p 5 + x 2- 3) ( p 5 + x 2 + 3) ( p 5 + x 2 + 3) = lim x ! 2 ( x- 2)( p 5 + x 2 + 3) 5 + x 2- 9 = lim x ! 2 ( x- 2)( p 5 + x 2 + 3) x 2- 4 = lim x ! 2 ( x- 2)( p 5 + x 2 + 3) ( x- 2)( x + 2) = lim x ! 2 p 5 + x 2 + 3 x + 2 = p 5 + 2 2 + 3 2 + 2 = 6 4 = 3 2 . (b) [6 points] lim x ! 3- 1- x 2 3- x . Solution: As x approaches 3, the numerator approaches- 8 and the denomina- tor approaches 0. Since x is approaching 3 from the left, x < 3 and so 3- x > 0. This means that the denominator is always positive as it approaches 0 and the fraction is negative when x is close enough to 3. Therefore, lim x ! 3- 1- x 2 3- x =-1 . (c) [6 points] lim x !-1 p x 2 + 1 x + 1 Solution: Note that when x is negative, x =- | x | =- p x 2 . So lim x !-1 p x 2 + 1 x + 1 = lim x !-1 p x 2 + 1 /x ( x + 1) /x = lim x !-1- p x 2 + 1 / p x 2 1 + 1 /x = lim x !-1- p ( x 2 + 1) /x 2 1 + 1 /x = lim x !-1- p 1 + 1 /x 2 1 + 1 /x . But 1 /x and 1 /x 2 both approach 0 as x approaches-1 , so lim x !-1 p x 2 + 1 x + 1 = lim x !-1- p 1 + 1 /x 2 1 + 1 /x =- p 1 + 0 1 + 0 =- 1 . 4. [15 points] Sketch a graph of y = ( x- 1) x 2 (2 x- 1) 2 ( x + 1) . Find and label the horizontal and vertical asymptotes and all intercepts....
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| s(3 s(1 | = | 2 | | 18 2 | = 22 meters 3 Evaluate the...

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