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Unformatted text preview: h )f ( x ) h = lim h → 1 x + h +11 x +1 h = lim h → x +1( x + h +1) ( x + h +1)( x +1) h = lim h →h h ( x + h +1)( x +1) = lim h →1 ( x + h +1)( x +1) =1 ( x +1) 2 . If the slope of the tangent line at a point x = a is1 9 , we must have f ( a ) =1 9 . Therefore,1 ( a +1) 2 =1 9 . Solving the equation for a , we ﬁnd two answers: a = 2 ,4. So we have two points on the graph of f with the desired property: (2 , 1 3 ) and (4 ,1 3 ). 4. (a) f ( t ) = e t . (b) g ( x ) = 1 √ x3 x4 . 1 MAC 2311 Sample; Solutions 2 5. (a) lim x → 3 f ( x ) = 2 lim x →f ( x ) = 2 (b) Vertical asymptotes: x =2. Horizontal asymptotes: y = 1 and y = 2. (c) a = 2 and a = 5 (d) a =2, a = 0, and a = 3....
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 Spring '08
 heister
 Limit of a function, 1. x1 x1, x1 x1 Left

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