MAC2311-SampleTest1Solutions

# MAC2311-SampleTest1Solutions - h-f x h = lim h → 1 x h...

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MAC 2311 Sample Test 1 solutions ————————————————– 1. (a) lim t 0 t +4 - 2 t = lim t 0 t +4 - 2 t · t +4+2 t +4+2 = lim t 0 ( t +4) - 4 t ( t +4+2) = lim t 0 1 t +4+2 = 1 4 (b) Right limit: lim x 1 + | x - 1 | x - 1 = lim x 1 + x - 1 x - 1 = 1. Left limit: lim x 1 - | x - 1 | x - 1 = lim x 1 - - ( x - 1) x - 1 = - 1. Since right-limit and left-limit are not equal, the limit does not exist. (c) In the limit lim x →∞ sin x 2 x 2 +1 the numerator always remains between -1 and 1 while the denominator tends to inﬁnity, so the limit is zero. 2. When x < 1, f ( x ) is given by x 2 + 1 which is continuous. Also, when x > 1, f ( x ) is given by 4 - x which is continuous. So, f ( x ) is continuous at every number x , except possibly at x = 1. At x = 1, f ( x ) does not have a limit because the left-limit and the right-limit are not equal: Left limit: lim x 1 - f ( x ) = lim x 1 - x 2 + 1 = 2; Right limit: lim x 1 + f ( x ) = lim x 1 + (4 - x ) = 3. So, x = 1 is the only point of discontinuity of f . 3. f 0 ( x ) = lim h 0 f ( x +

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Unformatted text preview: h )-f ( x ) h = lim h → 1 x + h +1-1 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 is-1 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 √ x-3 x-4 . 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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MAC2311-SampleTest1Solutions - h-f x h = lim h → 1 x h...

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