Problem Set 02s

Lim x 1 solution 4 3h h 2 4 lim h 0 h h rationalizing

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 2 (2 − x )(2 + x ) 2 + x 4 Solution: lim− 36 − x 2 = h →6 = lim− (36 − x 2 ) h →6 lim 36 − lim− x 2 h →6 − h →6 = 36 − (6)2 = 0 2 c. lim (2 + h ) −h h h →0 Solution: (2 + h )2 − h = lim 4 + 4h + h 2 − h lim h →0 h 2 4 − 3h + h = lim h →0 h h →0 h So lim f ( x) ≠ lim− f ( x) x →2+ h →0 e. lim x →1+ Solution: 4 − 3h + h 2 4 ≈ lim− = −∞ h →0 h h Rationalizing the denominator 4 − 3h + h 2 4 lim+ ≈ lim+ = ∞ h →0 h →0 h h = lim+ x →1 Therefore the limit does not exist. x →2 x2 −1 x 2 x −1 ⋅ x −1 x −1 = (x + 1)(x − 1) x − 1 = (x + 1) x − 1 x 2 (x − 1) x2 = lim + x→1 2− x 4 − x 2 1 x2 x − 1 1 x2 −1 1− 2 2 x = x2 = x −1 lim x →1+ x −1 x − 1 x 2 x − 1 From the right side d. lim 1− Since this is exponential function as h → 0 , we see that the denominator will approaches 4 and the numerator approaches zero. Therefore, from...
View Full Document

This document was uploaded on 02/27/2014 for the course M 408c at University of Texas.

Ask a homework question - tutors are online