exam2_2009_sol

# 3 evaluate the following limits or explain why they

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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 .

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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. Solution: The numerator is zero when x = 0 , 1, while the denominator is zero when x = - 1 , 1 / 2. This tells us that the x -intercepts (zeros) are x = 0 , 1 and the vertical asymptotes are at x = - 1 , 1 / 2.
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