This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
This note was uploaded on 10/24/2009 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.
 Fall '08
 SPEZIALE
 Math, Calculus

Click to edit the document details