MATH100problemSet1_s - Math 100 Section 105 Solutions to Problem Set 1 1(Calculation of limits(a For x 6 = 1 we have x 2 x x 1 = x x 1 x 1 = x and

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 100, Section 105 Solutions to Problem Set 1 September 27, 2010 1. (Calculation of limits) (a) For x 6 = 1 we have x 2- x x- 1 = x ( x- 1) x- 1 = x , and we know lim x → a x = a so lim x → 1 x 2- x x- 1 = 1 . (b) lim x → 1 x 2 + x x- 1 does not exist, since x- 1 → while the numerator has a non-zero limit. Alternatively, if lim x → 1 x 2 + x x- 1 existed and equalled L then the product rule would give 2 = lim x → 1 ( x 2 + x ) = lim x → 1 x 2 + x x- 1 · ( x- 1) = lim x → 1 x 2 + x x- 1 lim x → 1 ( x- 1) = L · 0 = 0 , a contradiction. (c) We have 1 h 1 (2+ h ) 2- 1 4 = 4- (2+ h ) 2 h (2+ h ) 2 4 =- 4 h + h 2 4 h (2+ h ) 2 =- 4+ h 4(2+ h ) 2 . Now by the sum rule, lim h → (4+ h ) = 4 + 0 = 4 and by the product rule lim h → 4(2 + h ) 2 = 4 lim h → (2 + h ) 2 = 4 · 2 2 = 16 . Since the last number is non-zero we may apply the quotient rule and nd lim h → 1 h 1 (2+ h ) 2- 1 4 =- 4 16 =- 1 4 ....
View Full Document

This note was uploaded on 11/16/2010 for the course MATH 100 math 100 taught by Professor Lior during the Spring '10 term at The University of British Columbia.

Page1 / 2

MATH100problemSet1_s - Math 100 Section 105 Solutions to Problem Set 1 1(Calculation of limits(a For x 6 = 1 we have x 2 x x 1 = x x 1 x 1 = x and

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online