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MIT18_014F10_pset4sols

# MIT18_014F10_pset4sols - 18.014 Problem Set 4 Solutions...

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18.014 Problem Set 4 Solutions Total: 24 points Problem 1: Establish the following limit formulas. You may assume the formula lim x 0 sin( x x ) = 1. (a) sin(5 x ) lim = 5 . x 0 sin( x ) (b) sin(5 x ) sin(3 x ) lim = 2 . x 0 x (c) 2 1 1 1 x lim = . x 0 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 0 sin( x ) x 0 5 x sin( x ) x 0 5 x · x 0 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 0 sin( x ) = 5 lim x 0 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 0 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 0 x x 0 x x 0 x Next, for any real number a = 0, we observe sin( ax ) sin( ax ) sin( x ) lim = a lim = a lim = a. x 0 x x 0 ax x 0 x 1

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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 0 x x 0 x x 0 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 0 x 2 x 0 x 2 x 0 (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 = 0 to get that the limit of the second term is 1 / 2. Multiplying the two terms together, the limit is 1 1 / 2 = 1 / 2. · x Problem 2: Let A ( x ) = f ( t ) dt where f ( t ) = 1 when t < 0 and f ( t ) = 1 when 2 t 0. Graph y = A ( x ) when x [ 2 , 2]. Using and δ ,
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