68 3.2 Limit Theorems Proof: Let us frst prove (i). Let {xn} be a sequence in D that converges to x0 and xn ̸ = x0 for every n. By Theorem 3.1.2, lim f (xn)= ℓ and lim g(xn)= m. n→∞ n→∞ It follows from Theorem 2.2.1 that lim ( f + g)(xn)= lim( f (xn)+ g(xn)) = ℓ + m. n→∞ n→∞ Since {xn} was an arbitrary sequence in D converging to x0 and xn ̸ = 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 m ̸ = 0, then x0 is a limit point of De. If x0 is a limit point of D, there is a sequence {un} in D converging to x0 such that un ≠ x0 for every n. Since m ≠ 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 ∈ De whenever 0 < |x − x0| < δ ,x ∈ D. Then un ∈ De for all n suffciently large, and hence x0 is a limit point of De. The rest of the proof of (iv) can be completed easily following the proof of (i). □ 2x + 2x − 3 ■ Example 3.2.1 Consider f : R \ {−7}→ R given by f (x)= . Then, combining all x + 7 parts of Theorem 3.2.1, we get limx →−2(x2 + 2x − 3) limx→−2 x2 + limx→−2 2x − limx →−2 3 lim f (x)= = x→−2 limx →−2(x + 7) limx →−2 x + limx →−2 7 (−2)2 (limx→−2 x)2 + 2limx→−2 x − limx →−2 3 + 2(−2) − 3 = = limx →−2 x+ limx →−2 7 −2+ 7 3 = − . 5 ■ Example 3.2.2 We proceed in the same way to compute the following limit: 1 +(2x − 1)2 limx→0 1 + limx →0(2x − 1)2 1+ 1 2 lim = = = . x→0 x2 + 7 limx→0 x2 + limx→0 7 0+ 7 7 ■ Example 3.2.3 We now consider 2x + 6x + 5 lim . x→−1 x + 1 Since the limit of the denominator is 0, we cannot apply directly part (iv) of Theorem 3.2.1. Instead, we frst simplify the expression keeping in mind that in the defnition of limit we never need to evaluate the expression at the limit point itself. In this case, this means we may assume that x ̸ = −1. For any such x we have x2 + 6x + 5 (x + 1)(x + 5) = = x + 5. x + 1 x + 1
RkJQdWJsaXNoZXIy NTc4NTAz