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