MATH
275_exam2_solution

X 2 9 dx 9 x x 3 3 3 2 x 3 3 9 x 4 3 27 27 3 18 8 3

• Notes
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( x 2 - 9) dx = [9 x - x 3 / 3] 3 - 2 + [ x 3 / 3 - 9 x ] 4 3 = ((27 - 27 / 3) - ( - 18 - ( - 8) / 3)) + ((64 / 3 - 36) - (27 / 3 - 27)) = 110 / 3 (b) Z 2 π/ 3 - π/ 2 cos(2 t ) dt = 1 2 Z 4 π/ 3 - π cos( t ) dt = [ 1 2 sin( t )] 4 π/ 3 - π = 1 2 (sin(4 π/ 3) - sin( π )) = 1 2 ( - 3 / 2 - 0) = - 3 4 In the first step we used the expansion-contraction properties of the integral. 1

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3. Give an ‘ ε - δ ’ proof of the statement lim x →- 1 x + 2 = 1. Fix ε > 0, we need to find δ > 0 so that 0 < | x + 1 | < δ implies | x + 2 - 1 | < ε . This inequality is equivalent to 1 - ε < x + 2 < 1 + ε . We may assume that ε < 1 and then the previous inequality is equivalent to (1 - ε ) 2 - 1 < x + 1 < (1 + ε ) 2 - 1 ε 2 - 2 ε < x + 1 < ε 2 + 2 ε ( * ) Choosing δ = 2 ε - ε 2 = ε (2 - ε ) (which is positive if ε < 1) will get the job done. Indeed, if | x +1 | < δ then ε 2 - 2 ε < x +1 < 2 ε - ε 2 < ε 2 +2 ε which means that inequality (*) holds. But then | x + 2 - 1 | < ε holds as well which is what we needed. 4. Find the following limits. (You can use anything we proved in class about limits, but you need to explain your steps!) (a) lim x 0 sin(5 x ) 1 + x - 1 - x = lim x 0 sin(5 x )( 1 + x + 1 - x ) ( 1 + x - 1 - x )( 1 + x + 1 - x ) = lim x 0 sin(5 x )( 1 + x + 1 - x ) 2 x = lim x 0 sin(5 x ) 5 x ( 1 + x + 1 - x ) 2 / 5 The function ( 1+ x + 1 - x ) 2 / 5 is continuous at zero so its limit is ( 1+0+ 1 - 0) 2 / 5 = 5. The limit of sin(5 x ) 5 x is the same as sin( x ) x as x 0 which is 1. Using the fact that the limit of products is the product of limits we get that lim x 0 sin(5 x ) 1 + x - 1
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