Solution-quiz7

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Memorial University of Newfoundland **We aren't endorsed by this school
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MATH 3000
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Mathematics
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Oct 20, 2023
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Math 3000 Fall 2020 Quiz 7 1. (5 marks) Find the following limit and use the definition to show your calculation: lim x 1 x ( x 2 - 1) x - 1 =? . ANS: First of all, the domain of this function is x - 1 6 = 0 which leads to x 6 = 1 as its domain. We let D = ( -∞ , 1) (1 , ). Now let c = 1, then let δ 1 = 1, such that (1 - δ 1 , 1 + δ 1 ) \ { 1 } = (0 , 2) \ { 1 } ⊂ D. Note, one can calculate the limit as lim x 1 x ( x 2 - 1) x - 1 = lim x 1 x ( x + 1) = 2 . Let > 0. Consider the inequality | f ( x ) - L | = x ( x 2 - 1) x - 1 - 2 = | x ( x + 1) - 2 | = | ( x - 1)( x + 2) | = | x - 1 | · | x + 2 | < . For all x (0 , 2) \ { 1 } , one has, | x + 2 | ≤ 4 = M. Therefore, for all x (0 , 2) \ { 1 } , | f ( x ) - L | ≤ M | x - 1 | < ⇐⇒ | x - 1 | ≤ M . Let δ = min { δ 1 , /M } > 0. In conclusion, for all > 0, one can find δ = min { δ 1 , /M } > 0, such that, for all x with 0 < | x - 1 | < δ , one gets | f ( x ) - L | ≤ M | x - 1 | < , and hence lim x 1 x ( x 2 - 1) x - 1 = 2 . 2. (5 marks) Find the following limit and use the definition to show your calculation: lim x + h p x 2 + 1 - x i =? ANS: First of all, it is easy to see that the domain of this function is the whole real numbers. We let D = ( -∞ , ). Note, one can calculate the limit as lim x + h p x 2 + 1 - x i = lim x + x 2 + 1 - x x 2 + 1 + x x 2 + 1 + x = lim x + 1 x 2 + 1 + x = 0 . Let > 0. Consider the inequality | f ( x ) - L | = p x 2 + 1 - x = 1 x 2 + 1 + x < . 1
Math 3000 Fall 2020 Quiz 7 Note that for all x > 0, one has x 2 + 1 > x 2 = x . Therefore, one has 1 x 2 + 1 + x 1 2 x 1 x = 1 x . We now let 1 x < which implies that x > 1 / . Now let G ( ) = 1 / . That is, for all > 0, one can find G ( ) = 1 / , such that for all x > G ( ), one has | f ( x ) - L | < and hence lim x + h p x 2 + 1 - x i = lim x + 1 x 2 + 1 + x = 0 . 3. (4 marks) Find and classify the discontinuity of the following function f ( x ) = 1 - x 1 - x 3 . ANS: First of all, the domain of this function is 1 - x 3 6 = 0 which leads to x 6 = 1 as its domain. We let D = ( -∞ , 1) (1 , ). So we check the discontinuity of point x = 1. Note that L 1 = lim x 1 - f ( x ) = lim x 1 - 1 - x 1 - x 3 = lim x 1 - 1 - x (1 - x )(1 + x + x 2 ) = lim x 1 - 1 1 + x + x 2 = 1 3 , L 2 = lim x 1 + f ( x ) = lim x 1 + 1 - x 1 - x 3 = lim x 1 + 1 - x (1 - x )(1 + x + x 2 ) = lim x 1 + 1 1 + x + x 2 = 1 3 . It follows that L = L 1 = L 2 = 1 3 , however, f ( x ) is undefined at x = 1. Hence x = 1 is the removable discontinuity. 4. (6 marks) Prove or disprove the uniform continuity of f ( x ) = x 3 - 6 x - 1 on I = (0 , 5). Prove that f ( x ) = 0 must have one real solution. ANS: For any > 0, for all x, y I , we consider the following inequality | f ( x ) - f ( y ) | = | x 3 - 6 x - 1 - y 3 + 6 y + 1 | = | x 3 - y 3 - 6( x - y ) | = | x - y || x 2 + xy + y 2 - 6 | < . In fact, for all x, y I , one has | x 2 + xy + y 2 - 6 | ≤ x 2 + | xy | + y 2 + 6 81 = M. We now let M | x - y | < ⇔ | x - y | < /M. Let δ = /M . Then, for any > 0, there is δ = /M , such that for all x, y I with | x - y | < δ , one has | f ( x ) - f ( y ) | = | x 3 - 6 x - 1 - y 3 + 6 y + 1 | = | x - y || x 2 + xy + y 2 - 6 | < . Hence f ( x ) = x 3 - 6 x - 1 is uniformly continuous on I = (0 , 5). Then we show f ( x ) = 0 must have one real solution. Since f ( x ) is uniformly continuous on I = (0 , 5), it follows that f ( x ) is continuous on I = (0 , 5). Note that lim x 0 f ( x ) = f (0) = - 1 < 0 and lim x 5 f ( x ) = f (5) = 94 > 0 . One has f ( x ) is continuous on I 0 = [0 , 5]. By Intermediate Value Theorem, there exists c I 0 = [0 , 5] such that f ( c ) = 0. Hence f ( x ) = 0 must have one real solution. 2
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