72 3.2 Limit Theorems q 1 1 Note that f (x)= 2 > M is equivalent to M 1 q > x2 since M > 0. It follows that M > |x|. x 1 2 Now, if we choose δ such that 0 < δ < M . Then, if 0 < |x| < δ we get 0 < x < δ 2 and δ 2 1 < M . Therefore 1 1 1 = M, x2 > δ 2 > 1 M as desired. Defnition 3.2.6 (limits at infnity) Let f : D → R. (i) When D is not bounded above, we say that f has limit ℓ as x → ∞ if for every ε > 0, there exists M ∈ R such | f (x) − ℓ| < ε for all x > M,x ∈ D and write limx →∞ f (x)= ℓ, (ii) When D is not bounded below, we say that f has limit ℓ as x →−∞ if for every ε > 0, there exists L ∈ R such | f (x) − ℓ| < ε for all x < L,x ∈ D and write limx →−∞ f (x)= ℓ. We can also defne limx →∞ f (x)= ±∞ and limx →−∞ f (x)= ±∞ in a similar way. ■ Example 3.2.8 We prove from the defnition that 3x2 3 lim = . x→∞ 2x2 + 1 2 The approach is similar to that for sequences, with the difference that x need not be an integer. Let ε > 0. We want to fnd M such that for all x > M, 3x2 3 − < ε. (3.3) 2x2 + 1 2 Now, 2 2 − 6x 3x 3 6x 2 − 3 −3 3 − = = = . 2x2 + 1 2(2x2 + 1) 4x2 + 2 2 4x2 + 2 To simplify the calculations it will be convenient to assume that x > 0. This assumption is justifed since we can always choose M > 0. 2 Since 4x + 2 > 4x2, we have 3x2 3 3 3 − = < (3.4) 2x2 + 1 4x2 + 2 4x2 2
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