73 We now find Mby solving the inequality 3 4x2 <ε for the variable x. We obtain 3 4ε <x2 or equivalently r 3 4ε <|x|. This inequality suggests that choosingM=q3 4ε will suffice. If x >M, then x >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 any n∈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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