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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 137 Fall 2011 Assignment 3 - Solutions 1. Use the , denition of limits to prove that lim x sin x = 0. Hint. | sin x | x for all x . Take any > 0. We are required to nd such that | sin x-| < whenever 0 < | x-| < ; that is, | sin x | < whenever 0 < | x | < . Using the hint, | sin x | x | x | , which we can set < . Therefore, choosing = , well have < | x | < = | sin x | < , and by denition 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 to be the number min { 1 , / 17 } . In other words, take to be either / 17 or 1, whichever is less. Assuming | x-2 | < , prove that | f ( x )-10 | < . Take any > 0. Let = min { 1 , / 17 } . Then both 0 < 1 and 0 < 17 are true....
View Full Document

Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online