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m147A3sol

# m147A3sol - MATH 147 Assignment#3 Solutions 1 A function is...

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MATH 147 Assignment #3 Solutions 1) A function is defined by f ( x ) = x + 5 a if x ≤ - 3 - ax + b if - 3 < x < 3 2 x + 10 b if x 3 where a and b are constants. Find values of a and b that will ensure that f is continuous for all x . Sketch the graph of the resulting function. Solution It is clear that f ( x ) is continuous except possibly at x = ± 3. Observe that lim x →- 3 - f ( x ) = lim x →- 3 - x + 5 a = - 3 + 5 a = f ( - 3) It follows that f ( x ) will be continuous at x = - 3 if and only if lim x →- 3 + f ( x ) = - 3 + 5 a . However, lim x →- 3 + f ( x ) = lim x →- 3 - - ax + b = 3 a + b Therefore f ( x ) is continuous at x = - 3 if and only if - 3 + 5 a = 3 a + b or b = 2 a - 3 . We also know that lim x 3 - f ( x ) = lim x 3 - - ax + b = - 3 a + b 1

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and that lim x 3 + f ( x ) = lim x 3 + 2 x + 10 b = 6 + 10 b = f (3) Therefore f ( x ) is continuous at x = 3 if and only if - 3 a + b = 6 + 10 b or - 3 a - 6 = 9 b. Substituting b = 2 a - 3 into the above equation leads to - 3 a - 6 = 18 a - 27 or a = 1 . This means that b = 2(1) - 3 = - 1. We have shown that f ( x ) is continuous if and only if a = 1 and b = - 1. 2) Evaluate the following limits if they exist: i) lim x 0 sin | x | x ii) lim x 1 | x |-| x - 2 | x - 1 iii) lim x 0 sin(4 x ) tan( πx ) iv)lim x 0 cos( 1 x ) Solutions: i) Observe that sin | x | x = ( sin( x ) x if x 0 sin( - x ) x if x < 0 . 2
From this it follows that lim x 0 - sin | x | x = lim x 0 - sin( - x ) x = lim x 0 - - sin( x ) x = - 1 while lim x 0 + sin | x | x = lim x 0 + sin( x ) x = lim x 0 + sin( x ) x = 1 Since the two one sided limits are different, the limit does not exist. ii) First observe that on the interval [0 , 2] the expression | x | - | x - 2 | becomes x - ( - ( x - 2)) = 2 x - 2. Therefore, for x [0 , 2], x 6 = 1 f ( x ) = 2 x - 2 x - 1 = 2. This observation shows that lim x 1 | x |-| x - 2 | x - 1 = 2. iii) Using the fundamental trig limit and the fact that lim θ 0 θ tan( θ ) = 1 we get that lim x 0 sin(4 x ) tan( πx ) = lim x 0 sin(4 x ) 4 x lim x 0 πx tan( πx ) 4 π = 4 π iv) Let x k = 1 2 and y k = 1 (2 k +1) π . Then x k 0 and y k 0, but 3

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lim k →∞ cos( 1 x k ) = lim k →∞ cos(2 ) = 1 while lim k →∞ cos( 1 y k ) = lim k →∞ cos((2 k + 1) π
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