Problem Set 02s

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

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

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