62 3.1 Limits of Functions (c) Let δ >0 and let D= (4−δ,4+δ)\ {4}= (4−δ,4)∪(4,4+δ). Then 4 is a limit point of D. (d) Let D=Q. Then every real number is a limit point of D. This follows directly from the density of Q(see Theorem 1.6.3 and also Exercise 2.1.8). Definition 3.1.3 Let f : D→Rand let x0 be a limit point of D. We say that f has a limit at x0 if there exists a real number ℓ such that for any real number ε >0, there exists δ >0 for which | f (x)−ℓ| <ε for all x ∈Dwith 0<|x−x0| <δ. In this case, we say that ℓ is the limit of the function f at x0 and we write lim x→x0 f (x) =ℓ. Remark 3.1.1 If the limit of a function at a point exists, it is unique. The proof of this fact is similar to the one given for sequences in Theorem 2.1.1 and we leave it as an exercise. Remark 3.1.2 Note that the limit point x0 in the definition of limit of a function may or may not be an element of the domain D. In any case, the inequality | f (x)−ℓ| <ε needs only be satisfied by elements x of D. ■ Example 3.1.2 Let f : R→Rbe given by f (x) =5x−7. We will prove that limx→2 f (x) =3. First notice that in this example, x0 =2 andℓ =3. Given anyε >0, we seekδ >0 such that if x∈R with 0<|x−2| <δ, then| f (x)−3| <ε. To find a suitable δ, we start by simplifying the expression | f (x)−3|. We have | f (x)−3| =|(5x−7)−3| =|5x−10| =|5(x−2)| =|5||x−2| =5|x−2|. Now we needδ >0 such that if 0<|x−2| <δ, then 5|x−2| <ε. This suggests that δ =ε/5 is a good choice since then 5|x−2| <5δ =5(ε/5) =ε. Now that we have found δ, the last step is to write a formal proof. Let ε >0 and consider δ =ε/5>0. If x ∈Rand 0<|x−2| <δ, we have | f (x)−3| =5|x−2| <5δ =ε. This shows that limx→2 f (x) =3. ■ Example 3.1.3 Let f : [0,1) →Rbe given by f (x) =x 2+7x. We will show that limx→1 f (x) =8. In this case, x0 =1 and ℓ =8. Following the previous example, let us start by simplifying the expression| f (x)−ℓ|. We have | f (x)−ℓ| =|(x2 +7x)−8| =|x2 +7x−8| =|x+8||x−1|. Since D= [0,1), anyx ∈Dsatisfies the inequality |x| <1. This allows us to find an upper bound for |x+8| using the triangle inequality. If x ∈[0,1), then |x+8| ≤ |x| +8<1+8=9.
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