63 Now, given ε >0, we need to find a suitable δ >0 such that if x ∈[0,1) with 0<|x−1| <δ, then | f (x)−8| <ε. We have | f (x)−ℓ| =|(x2 +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 | f (x)−8| =|x+8||x−1| <9δ =9 ε 9 =ε. This shows that limx→1 f (x) =8. ■ Example 3.1.4 Let f : R→Rbe 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=Ris not a bounded domain. We cannot proceed in the same way as in Example 3.1.3. The simplification 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1+8=9 is no longer valid.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 suchx, that |x| <2 and, hence |x+8| ≤ |x| +8<2+8=10. Now, givenε >0 we choose δ =min{1, ε 10}. Then, whenever |x−1| <δ we get | f (x)−8| =|x+8||x−1| ≤(|x| +8)|x−1| <10δ ≤10 ε 10 =ε. This shows that limx→1 f (x) =8. ■ Example 3.1.5 Let f : R→Rbe given by f (x) = 3x−5 x2 +3 . We prove that limx→1 f (x) =−1/2. First we look at the expression | f (x)−(−1 2)| and try to identify a factor |x−1| (because here x0 =1). We have f (x)− − 1 2 = 3x−5 x2 +3 + 1 2 = 6x−10+x2 +3 2(x2 +3) = x2 +6x−7 2(x2 +3) =| x−1||x+7| |2(x2 +3)| . Since|2(x2+3)| =2(x2+3), 2(x2+3)≥x2+3 andx2+3≥3 for all x∈R, we obtain the following estimate: |x−1||x+7| |2(x2 +3)| =| x−1||x+7| 2(x2 +3) ≤ | x−1||x+7| x2 +3 ≤ 1 3| x−1||x+7|. 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. Now, given ε >0, we choose δ =min{1, ε 3}. It follows that if x ∈Rwith 0<|x−1| <δ, then f (x)−(− 1 2 ) ≤ | x+7| 3 | x−1| < 9 3 δ =3δ ≤ε. We have proved that limx→1 f (x) =−1/2.
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