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Unformatted text preview: MATH 137 Assignment 3 Solutions to Problems 110 Question 1. It can be shown that sin x x 1 as x . Use this fact to show that 1 cos x x as x . Solution. We write 1 cos x x = (1 cos x )(1 + cos x ) x (1 + cos x ) = 1 cos 2 x x (1 + cos x ) = sin 2 x x (1 + cos x ) (by using that sin 2 x + cos 2 x = 1 ) = sin x x sin x 1 + cos x . We were reminded in the statement of the problem that lim x sin x x = 1 , (1) and from the continuity of sin and cos it follows that lim x sin x 1 + cos x = sin 0 1 + cos 0 = 1 + 1 = 0 . (2) By doing the product of the limits (1) and (2) it thus follows that lim x 1 cos x x = lim x sin x x lim x sin x 1 + cos x = 1 0 = 0 . Another Solution to Question 1. We use the formula for the cosine of the double angle, applied to the situation of x = (double of x/ 2 ). This gives cos x = 1 2 sin 2 ( x/ 2) , hence that 1 cos x = 2 sin 2 ( x/ 2) . So then 1 cos x x = 2 sin 2 ( x/ 2) x = sin( x/ 2) x/ 2 sin( x/ 2) . 1 When x we have that x/ 2 , which implies in turn that lim x sin( x/ 2) x/ 2 = 1 , lim x sin( x/ 2) = sin 0 = 0 . By doing a product of limits we thus find that lim x 1 cos x x = lim x sin( x/ 2) x/ 2 lim x sin( x/ 2) = 1 0 = 0 . Question 2. Use the limit laws along with the basic limit lim x sin x x = 1 to evaluate lim x x 2 (3 + sin x ) ( x + sin x ) 2 . Solution. We observe first that lim x x + sin x x = lim x 1 + sin x x = 1 + lim x sin x x = 2 . By taking the reciprocal of this limit, it thus follows that lim x x x + sin x = 1 / 2 . Now the limit given in the question can be written as lim x x x + sin x 2 (3 + sin x ) , By invoking the continuity of the sine function and the rules of operations with limits, we therefore find that the required limit is equal to (1 / 2) 2 (3+sin 0) = 3 / 4 . Question 3. With a bit of thinking, evaluate lim x x  2 x 1    2 x + 1  . Solution. Consider a number x which is close to 0, for instance x ( 1 / 10 , +1 / 10) . Then 2 x 1 ( 6 / 5 , 4 / 5) , and in particular we see that 2 x 1 < . So for x close to 0, we have that  2 x 1  = (2 x 1) = 1 2 x . 2 Likewise, for x sufficiently close to we find that 2 x + 1 is positive, and hence that  2 x + 1  = 2 x + 1 . So then for x sufficiently close to 0 (e.g. for x ( 1 / 10 , 1 / 10) ), the expression given in this question is equal to x (1 2 x ) (1 + 2 x ) = x 4 x = 1 / 4 . It follows that, when x , the limit of the given expression is 1 / 4 . Question 4. Find lim x arctan x cos(1 /x ) , and justify your method. Solution. By writing  arctan x cos(1 /x )  =  arctan x   cos(1 /x )  and by taking into account that  cos(1 /x )  1 , we find that  arctan x cos(1 /x )   arctan x  , for every x R , x 6 = 0 . (3) Now the function  arctan x  is continuous on R , since it is obtained by compos ing the absolute value function with arctan , where both the absolute value and the...
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 Fall '08
 SPEZIALE
 Math, Calculus

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