Take f, a and ϕ as in theorem. Show with the calculation rules that: a. for every k ∈ R holds limx→a kf(x) = kϕ; b. if P(x) is a polynomial is then limx→a P(x) = P(a); c. for every k ∈ Q holds limx→a(f(x))k = ϕk, provided k > 0.
Take f, a and ϕ as in theorem. Show with the calculation rules that: a. for every k ∈ R holds limx→a kf(x) = kϕ; b. if P(x) is a polynomial is then limx→a P(x) = P(a); c. for every k ∈ Q holds limx→a(f(x))k = ϕk, provided k > 0.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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I have the following theorem: (Calculation rules for limits)
Let limx→a f(x) = ϕ and limx→a g(x) = γ. Then the following statements apply:
a. limx→a
(f(x) + g(x)) = ϕ + γ
b. limx→a
(f(x) − g(x)) = ϕ − γ
c. limx→a
(f(x) g(x)) = ϕ γ
d. limx→a f(x)/g(x)=ϕ/γ provided γ ≠ 0
e. limx→a f(g(x)) = limx→γ f(x).
Now I have the question that needs this theorem:
Take f, a and ϕ as in theorem. Show with the calculation rules that:
a. for every k ∈ R holds limx→a kf(x) = kϕ;
b. if P(x) is a polynomial
is then limx→a P(x) = P(a);
c. for every k ∈ Q holds limx→a(f(x))k = ϕk, provided k > 0.
Please do it step by step in detail, thank you in advance.
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