This preview shows pages 1–4. Sign up to view the full content.
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 147 Assignment #3 Solutions 1) A function is defined by f ( x ) = x + 5 a if x ≤  3 ax + b if 3 < x < 3 2 x + 10 b if x ≥ 3 where a and b are constants. Find values of a and b that will ensure that f is continuous for all x . Sketch the graph of the resulting function. Solution It is clear that f ( x ) is continuous except possibly at x = ± 3. Observe that lim x → 3 f ( x ) = lim x → 3 x + 5 a = 3 + 5 a = f ( 3) It follows that f ( x ) will be continuous at x = 3 if and only if lim x → 3 + f ( x ) = 3 + 5 a . However, lim x → 3 + f ( x ) = lim x → 3 ax + b = 3 a + b Therefore f ( x ) is continuous at x = 3 if and only if 3 + 5 a = 3 a + b or b = 2 a 3 . We also know that lim x → 3 f ( x ) = lim x → 3 ax + b = 3 a + b 1 and that lim x → 3 + f ( x ) = lim x → 3 + 2 x + 10 b = 6 + 10 b = f (3) Therefore f ( x ) is continuous at x = 3 if and only if 3 a + b = 6 + 10 b or 3 a 6 = 9 b. Substituting b = 2 a 3 into the above equation leads to 3 a 6 = 18 a 27 or a = 1 . This means that b = 2(1) 3 = 1. We have shown that f ( x ) is continuous if and only if a = 1 and b = 1. 2) Evaluate the following limits if they exist: i) lim x → sin  x  x ii) lim x → 1  x  x 2  x 1 iii) lim x → sin(4 x ) tan( πx ) iv)lim x → cos( 1 x ) Solutions: i) Observe that sin  x  x = ( sin( x ) x if x ≥ sin( x ) x if x < . 2 From this it follows that lim x → sin  x  x = lim x → sin( x ) x = lim x → sin( x ) x = 1 while lim x → + sin  x  x = lim x → + sin( x ) x = lim x → + sin( x ) x = 1 Since the two one sided limits are different, the limit does not exist. ii) First observe that on the interval [0 , 2] the expression  x    x 2  becomes x ( ( x 2)) = 2 x 2. Therefore, for x ∈ [0 , 2], x 6 = 1 f ( x ) = 2 x 2 x 1 = 2. This observation shows that lim x → 1  x  x 2  x 1 = 2....
View
Full
Document
This note was uploaded on 10/21/2010 for the course MATH 147 taught by Professor Wolzcuk during the Fall '09 term at Waterloo.
 Fall '09
 Wolzcuk
 Math

Click to edit the document details