68 3.2 Limit Theorems Proof: Let us first prove (i). Let {xn}be a sequence in Dthat converges tox0 andxn̸ =x0 for every n. By Theorem 3.1.2, lim n→∞ f (xn) =ℓ and lim n→∞ g(xn) =m. It follows from Theorem 2.2.1 that lim n→∞ (f +g)(xn) =lim n→∞ (f (xn)+g(xn)) =ℓ+m. Since {xn} was an arbitrary sequence in Dconverging to x0 and xn̸ =x0 for every n, applying Theorem 3.1.2 again, we get limx→x0(f +g)(x) =ℓ+m. The proofs of (ii) and (iii) are similar. We will show that if m̸=0, then x0 is a limit point of e D. If x0 is a limit point of D, there is a sequence {un} in Dconverging to x0 such that un̸ =x0 for every n. Since m̸=0, it follows from Theorem 3.1.6 that there exists δ >0 with g(x)̸ =0 whenever 0<|x−x0| <δ,x ∈D. This implies x ∈ e Dwhenever 0<|x−x0| <δ,x ∈D. Then un ∈ e Dfor all n sufficiently large, and hence x0 is a limit point of e D. The rest of the proof of (iv) can be completed easily following the proof of (i). □ ■ Example 3.2.1 Consider f : R\ {−7} →Rgiven by f (x) = x2 +2x−3 x+7 . Then, combining all parts of Theorem 3.2.1, we get lim x→−2 f (x) = limx→−2(x 2 +2x−3) limx→−2(x+7) = limx→−2x 2 +limx→−22x−limx→−23 limx→−2x+limx→−27 = (limx→−2x) 2 +2 limx→−2x−limx→−23 limx→−2x+limx→−27 = (−2)2 +2(−2)−3 −2+7 =− 3 5 . ■ Example 3.2.2 We proceed in the same way to compute the following limit: lim x→0 1+(2x−1)2 x2 +7 = limx→01+limx→0(2x−1) 2 limx→0x2 +limx →07 = 1+1 0+7 = 2 7 . ■ Example 3.2.3 We now consider lim x→−1 x2 +6x+5 x+1 . Since the limit of the denominator is 0, we cannot apply directly part (iv) of Theorem 3.2.1. Instead, we first simplify the expression keeping in mind that in the definition of limit we never need to evaluate the expression at the limit point itself. In this case, this means we may assume that x̸ =−1. For any such x we have x2 +6x+5 x+1 = (x+1)(x+5) x+1 =x+5.
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