Concept explainers
Finding a Limit In Exercises 71-76, find each limit.
(a)
(b)
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Calculus: Early Transcendental Functions
- " (Sum Rule): Suppose f: ℝⁿ → ℝᵐ and g: ℝⁿ → ℝᵐ are functions, and let a ∈ ℝⁿ and b, c ∈ ℝᵐ be points. If lim(x→a) f(x) = b and lim(x→a) g(x) = c, then lim(x→a) (f(x) + g(x)) = b + c. Proof: Assume that lim(x→a) f(x) = b and lim(x→a) g(x) = c. Let ε > 0 be arbitrary. Then there exists δ₁ > 0 such that for x ∈ Dom(f) with d(x,a) < δ₁, we have ||f(x) - b|| < ε/2 (Equation 1.9). Similarly, there exists δ₂ > 0 such that for x ∈ Dom(g) with d(x,a) < δ₂, we have ||g(x) - c|| < ε/2 (Equation 1.10). Take δ := min(δ₁, δ₂) and let x ∈ Dom(f + g) satisfy d(x,a) < δ. Since x ∈ Dom(f) and d(x,a) < δ₁, Equation 1.9 holds. Furthermore, x ∈ Dom(g) and d(x,a) < δ₂, so Equation 1.10 applies. We can combine these inequalities: ||f(x) + g(x) - (b + c)|| = ||(f(x) - b) + (g(x) - c)|| ≤ ||f(x) - b|| + ||g(x) - c|| < ε/2 + ε/2 = ε. This shows that for all x ∈ Dom(f + g) with d(x,a) < δ, we have ||f(x) + g(x) - (b + c)|| < ε. Therefore, f(x) + g(x) → b + c as x → a." I…arrow_forwardConsider the function f(x, y) 2y (a) Show that lim f(x, y) does not exist by using different paths (z,u)-(0,0) toward the origin. Provide a diagram to illustrate the paths that you have chosen. (b) Show that lim f(x, y) does not exist by using e – 6 definition. (1,y)-(0,0)arrow_forwardEvaluate the indicated limit or explain why it does not exists: x² (y - 1)² lim (x,y) (0,1) x² + (y − 1)²arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage