Homework2

# Homework2 - Ma 115 Homework Solutions for Week 2 2.2 #4 x0...

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2.2 # 4 ( a ) lim x 0 f ( x ) = 3 ( b ) lim x 3 - f ( x ) = 4 ( c ) lim x 3 + f ( x ) = 2 ( d ) lim x 3 f ( x ) does not exist because the limits in (b) and (c) are not equal. ( e ) f (3) = 3 2.2 # 6 lim x a f ( x ) exists for all a except a = ± 1 2.3 # 14 lim h 0 1+ h - 1 h = lim h 0 1+ h - 1 h · 1+ h +1 1+ h +1 = lim h 0 (1+ h ) - 1 h ( 1+ h +1 ) = lim h 0 h h ( 1+ h +1 ) = lim h 0 1 1+ h +1 = 1 1+1 = 1 2 2.3 # 16 lim x 2 x 4 - 16 x - 2 = lim x 2 ( x +2)( x - 2) ( x 2 +4 ) x - 2 = lim x 2 ( x + 2)( x 2 + 4) = (2 + 2)(2 2 + 4) = 32 2.3 # 20 lim t 0 £ 1 t - 1 t 2 + t / = lim t 0 ( t 2 + t ) - t t ( t 2 + t ) = lim t 0 t 2 t 2 ( t +1) = lim t 0 1 t +1 = 1 0+1 = 1 2.3 # 28 - 1 sin( π / x ) 1 e - 1 e sin( π / x ) e 1 (only because e x is a monotone increasing function) x e xe sin( π / x ) x · e Since lim x 0 + x e · = 0 and lim

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## Homework2 - Ma 115 Homework Solutions for Week 2 2.2 #4 x0...

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