MIT18_014F10_pset4sols

MIT18_014F10_pset4sols - 18.014 Problem Set 4 Solutions...

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Unformatted text preview: 18.014 Problem Set 4 Solutions Total: 24 points Problem 1: Establish the following limit formulas. You may assume the formula lim x sin( x x ) = 1. (a) sin(5 x ) lim = 5 . x sin( x ) (b) sin(5 x ) sin(3 x ) lim = 2 . x x (c) 2 1 1 1 x lim = . x x 2 2 Solution (4 points) (a) Using the product formula for limits (Thm. 3.1 part iii), we have sin(5 x ) sin(5 x ) 5 x sin(5 x ) 5 x lim = lim = lim lim = AB. x sin( x ) x 5 x sin( x ) x 5 x x sin( x ) For the first term, note that 5 x approaches zero as x approaches zero; hence, 5 x A = 1 by the assumed limit formula. For the second term, note lim x sin( x ) = 5 lim x sin( 1 x ) /x = 5 1 = 5 by the product rule and the quotient rule (Thm. 3.1 part iv). Thus, B = 5 and sin(5 x ) lim = 5 x sin( x ) as desired. (b) Here we use the difference rule (Thm. 3.1 part ii) to obtain sin(5 x ) sin(3 x ) sin(5 x ) sin(3 x ) lim = lim lim . x x x x x x Next, for any real number a = 0, we observe sin( ax ) sin( ax ) sin( x ) lim = a lim = a lim = a. x x x ax x x 1 Plugging back into the above formula yields sin(5 x ) sin(3 x ) sin(5 x ) sin(3 x ) lim = lim lim = 5 3 = 2 . x x x x x x (c) We use the product rule to get 2 1 1 x (1 1 x 2 )(1 + 1 x 2 ) 1 lim = lim lim . x x 2 x x 2 x (1 + 1 x 2 ) Since (1 1 x 2 )(1 + 1 x 2 ) = 1 (1 x 2 ) = x 2 , the first limit is one. For the 1 2 second limit, note that is the composition of the functions 1 x , x , 1+ x , 2 1+ 1 x and x 1 , which are continuous at the points 0, 1, 1, and 2 by Example 5 and Theorem 3 . 2. Hence, by Theorem 3 . 5, the function 1+ 1 1 x 2 is continuous at x = 0, and we can just plug in x = to get that the limit of the second term is 1 / 2. Multiplying...
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This note was uploaded on 01/18/2012 for the course MATH 18.014 taught by Professor Christinebreiner during the Fall '10 term at MIT.

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MIT18_014F10_pset4sols - 18.014 Problem Set 4 Solutions...

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