assignment 3 solutions

# assignment 3 solutions - MATH 137 Assignment 3 Solutions to...

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Unformatted text preview: MATH 137 Assignment 3 Solutions to Problems 1-10 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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## This note was uploaded on 10/24/2009 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.

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assignment 3 solutions - MATH 137 Assignment 3 Solutions to...

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