# sol2 - Problem Set 2 Calculus 1 Abdullah Khalid Hassan...

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Problem Set 2 Calculus 1 Abdullah Khalid, Hassan Bukhari November 23, 2010 Exercise 2.3 9. We have f ( x ) = x,x o = 1 ,L = 1 = 1 4 . We have the graph given in the book. As can be seen L + ± = 1 + 1 4 = 5 4 and L - ± = 1 - 1 4 = 3 4 . Consider that when f ( x ) lies in the interval ( 3 4 , 5 4 ), x is in the interval ( 9 16 , 25 16 ), and x o = 1 belongs to this interval. Now the lenght of ( 9 16 , 1) is smaller than length of (1 , 25 16 ). So we let δ = 1 - 9 16 = 7 16 . We can also solve this algebrically. Let’s start with, | f ( x ) - L | < ± Plug in the values of everything and manipulate, | x - 1 | < 1 4 - 1 4 < x - 1 < 1 4 3 4 < x < 5 4 9 16 < x < 25 16 - 7 16 < x - 1 < 9 16 ⇒ | x - 1 | < 7 16 which is of the form, | x - x o | < 7 16 Therefore, we have δ = 7 16 . 10. We have f ( x ) = 2 x + 1 ,x o = 3 ,L = 4 = 0 . 2 = 1 5 . Again we can see that L + ± = 4 + 0 . 2 = 4 . 2 and L - ± = 1 - 0 . 2 = 3 . 8. Consider that 1

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when f(x) lies in the interval (3 . 8 , 4 . 2), x is in the interval (2 . 61 , 3 . 41), and x o = 3 belongs to this interval. Now the lenght of (2 . 61 , 1) is smaller than length of (3 , 3 . 41). So we let δ = 3 - 2 . 61 = 0 . 39. Solving algebrically, we proceed as before, | f ( x ) - L | < ± | 2 x + 1 - 4 | < 1 5 - 1 5 < 2 x + 1 - 4 < 1 5 - 19 5 < 2 x + 1 < 21 5 - 19 10 < x + 1 < 21 10 - 19 2 10 2 < x + 1 < 21 2 10 2 ⇒ - 0 . 39 < x - 3 < 0 . 41 ⇒ | x - 3 | < 0 . 39 which is of the form, | x - x o
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sol2 - Problem Set 2 Calculus 1 Abdullah Khalid Hassan...

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