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soln3 - MATH 137 Fall 2011 Assignment 3 Solutions 1 Use the...

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Unformatted text preview: MATH 137 Fall 2011 Assignment 3 - Solutions 1. Use the ε , δ definition of limits to prove that lim x → sin x = 0. Hint. | sin x | ≤ x for all x . Take any ε > 0. We are required to find δ such that | sin x-| < ε whenever 0 < | x-| < δ ; that is, | sin x | < ε whenever 0 < | x | < δ . Using the hint, | sin x | ≤ x ≤ | x | , which we can set < ε . Therefore, choosing δ = ε , we’ll have < | x | < δ = ⇒ | sin x | < ε , and by definition lim x → sin x = 0. 2. Let f ( x ) = x 3-2 x + 6. In order to prove that f ( x ) → 10 as x → 2, complete the following steps. (a) Show that f ( x )-10 = ( x-2)( x 2 + 2 x + 2). f ( x )-10 = x 3-2 x-4. x = 2 is a zero of the polynomial, so ( x-2) is a factor and by long division we can see that x 3-2 x-4 = ( x-2)( x 2 + 2 x + 2). (or just expand out the right side.) (b) If | x-2 | < 1, prove that | x 2 + 2 x + 2 | < 17. | x-2 | < 1 = ⇒ x ∈ (1 , 3), so | x | < 3. Using the triangle inequality, | x 2 + 2 x + 2 | ≤ | x 2 | + | 2 x | + | 2 | = | x | 2 + 2 | x | + 2 < 3 2 + 2(3) + 2 = 17 (c) If | x-2 | < 1, prove that | f ( x )-10 | < 17 | x-2 | . | f ( x )-10 | = | ( x-2)( x 2 + 2 x + 2) | = | x-2 | · | x 2 + 2 x + 2 | from (a) ≤ 17 | x-2 | from (b) Note there is equality when x = 2. (d) Now let ± be any positive number. For this ± , take δ to be the number min { 1 , ±/ 17 } . In other words, take δ to be either ±/ 17 or 1, whichever is less. Assuming | x-2 | < δ , prove that | f ( x )-10 | < ± . Take any ε > 0. Let δ = min { 1 , ±/ 17 } . Then both 0 < δ ≤ 1 and 0 < δ ≤ ε 17 are true....
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soln3 - MATH 137 Fall 2011 Assignment 3 Solutions 1 Use the...

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