Prove the Corollary by mathematical induction (Real Analysis) Suppose that lim f(x)=L and lim f(x)=L. x-a x-a

 Prove the Corollary by mathematical induction (Real Analysis)

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Suppose that lim f₁(x)=L₁ and lim f₂(x)=L₂. x-a x-a Define g(x) = f₁(x) + f₂(x). Then and lim g(x) L₁ + L₂. x-a PROOF. Let &> 0 be given. Then, using the quantity &/2, there are positive numbers d, and 8₂ such that |f₁(x) - L₁ <for all x satisfying 0 < x-a < ₁ |f₂(x) - L₂|<for all x satisfying 0 <|x − a| < 8₂. Defined as the smaller of 8₁ and 8₂. Then lg(x) (L₁ + L₂)| = |f₁(x) - L₁ + f₂(x) - L₂l <fi(x) - L₁ + f₂(x) - L₂l₂ and for 0 < x-al < 6, it follows that E & g(x) - (L₂ + L₂)| <+₁ <= E. The result is established. 0 Corollary. Suppose that limafi(x) = L₁, i= 1, 2, ..., n. Define g(x) = Ei=1 f(x). Then lim g(x) = [ L₁. x-a i=1 Suppose that lim f₁(x)=L₁ and lim f₂(x)=L₂. x-a x-a Define g(x) = f₁(x) + f₂(x). Then and lim g(x) L₁ + L₂. x-a PROOF. Let &> 0 be given. Then, using the quantity &/2, there are positive numbers d, and 8₂ such that |f₁(x) - L₁ <for all x satisfying 0 < x-a < ₁ |f₂(x) - L₂|<for all x satisfying 0 <|x − a| < 8₂. Defined as the smaller of 8₁ and 8₂. Then lg(x) (L₁ + L₂)| = |f₁(x) - L₁ + f₂(x) - L₂l <fi(x) - L₁ + f₂(x) - L₂l₂ and for 0 < x-al < 6, it follows that E & g(x) - (L₂ + L₂)| <+₁ <= E. The result is established. 0 Corollary. Suppose that limafi(x) = L₁, i= 1, 2, ..., n. Define g(x) = Ei=1 f(x). Then lim g(x) = [ L₁. x-a i=1

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