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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 approaches1 , 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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 Fall '07
 GOMEZ,JONES
 Math, Calculus, Algebra, Trigonometry, Derivative, Product Rule, lim

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