63 Now, given ε > 0, we need to fnd a suitable δ > 0 such that if x ∈ [0,1) with 0 < |x − 1| < δ , then | f (x) − 8| < ε. We have 2 | f (x) − ℓ| = |(x + 7x) − 8| = |x + 8||x − 1|≤ 9|x − 1|. Next we determine δ > 0 such that if 0 < |x − 1| < δ , then 9|x − 1| < ε. This suggests to select δ = ε/9. We now write a formal proof. Let ε > 0 and consider δ = ε/9 > 0. If x ∈ [0,1) is such that 0 < |x − 1| < δ , then ε | f (x) − 8| = |x + 8||x − 1| < 9δ = 9 = ε. 9 This shows that limx →1 f (x)= 8. ■ Example 3.1.4 Let f : R → R be given by f (x)= x 2 + 7x. We show that limx→1 f (x)= 8. We have x0 = 1 and ℓ = 8. Note that even though the expression of the function given is the same as in the previous example, the domain is not. In this example, D = R is not a bounded domain. We cannot proceed in the same way as in Example 3.1.3. The simplifcation of the expression | f (x) − ℓ| is the same as before. Since the domain is all of R, the estimate |x + 8|≤|x| + 8 < 1 + 8 = 9 is no longer valid. However, we are interested only in values of x close to x0 = 1. Thus, we impose the condition δ ≤ 1 (we can choose any positive number we like). If |x− 1| < 1, then −1 < x− 1 < 1, so 0 < x < 2. It follows, for such x, that |x| < 2 and, hence |x + 8|≤|x| + 8 < 2+ 8 = 10. ε Now, given ε > 0 we choose δ = min{1, 10 }. Then, whenever |x − 1| < δ we get ε | f (x) − 8| = |x + 8||x − 1|≤ (|x| + 8)|x − 1| < 10δ ≤ 10 = ε. 10 This shows that limx →1 f (x)= 8. 3x − 5 ■ Example 3.1.5 Let f : R → R be given by f (x)= . We prove that limx →1 f (x)= −1/2. x2 + 3 First we look at the expression | f (x) − (−1 2 )| and try to identify a factor |x − 1| (because here x0 = 1). We have 2 2 1 3x − 5 1 6x − 10 + x + 3 x + 6x − 7 |x − 1||x + 7| f (x) − − = x2 + 3 + = 2(x2 + 3) = 2(x2 + 3) = |2(x2 + 3)| . 2 2 2 2 2 2 2 Since |2(x +3)| = 2(x +3), 2(x +3) ≥ x +3 and x +3 ≥ 3 for all x ∈ R, we obtain the following estimate: |x − 1||x + 7| |x − 1||x + 7| |x − 1||x+ 7| 1 |2(x2 + 3)| = 2(x2 + 3) ≤ x2 + 3 ≤ | x − 1||x + 7|. 3 Proceeding as in the previous example, if |x − 1| < 1 we get −1 < x − 1 < 1, so 0 < x < 2. Thus, |x| < 2 and |x + 7|≤|x| + 7 < 9. ε Now, given ε > 0, we choose δ = min{1, 3 }. It follows that if x ∈ R with 0 < |x − 1| < δ , then 1 |x + 7| 9 f (x) − (− ) ≤ |x − 1| < δ = 3δ ≤ ε. 2 3 3 We have proved that limx →1 f (x)= −1/2.
RkJQdWJsaXNoZXIy NTc4NTAz