73 We now findMby solving the inequality 3 4x2 <ε for the variable x. We obtain 3 4ε <x2 or equivalentlyr 3 4ε <|x|. This inequality suggests that choosingM=q3 4ε will suffice. If x >M, thenx >q3 4ε andx 2 > 3 4ε . Hence, 3x2 2x2 +1− 3 2 = 3 4x2 +2 < 3 4x2 <ε. Therefore lim x→∞ 3x2 2x2 +1 = 3 2 . Exercises 3.2.1 Find the following limits: (a) lim x→2 3x2 −2x+5 x−3 , (b) lim x→−3 x2 +4x+3 x2 −9 . 3.2.2 Let f : D→Rand let x0 is a limit point of D. Prove that if limx →x0 f (x) exists, then lim x→x0 [ f (x)]n =[ lim x→x0 f (x)]n,for anyn∈N. 3.2.3 Find the following limits: (a) lim x→1 √x −1 x2 −1 , (b) lim x→1 xm−1 xn −1 , where m,n∈N, (c) lim x→1 n√x −1 m√x −1 , where m,n∈N, m,n≥2, (d) lim x→1 √x − 3 √x x−1 . 3.2.4 Find the following limits: (a) limx →∞( 3√x3 +3x2 − √x2 +1), (b) limx →−∞( 3√x3 +3x2 − √x2 +1).
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