Proof Prove that if the limit of f ( x ) as x approaches c exists, then the limit must be unique. [ Hint: Let lim x → 0 f ( x ) = L 1 , and lim x → 0 f ( x ) = L 2 , and prove that L 1 = L 2 .]
Proof Prove that if the limit of f ( x ) as x approaches c exists, then the limit must be unique. [ Hint: Let lim x → 0 f ( x ) = L 1 , and lim x → 0 f ( x ) = L 2 , and prove that L 1 = L 2 .]
Solution Summary: The author explains that the limit of a function f(x) as x approaches c must be unique.
Proof Prove that if the limit of f(x) as x approaches c exists, then the limit must be unique. [Hint: Let
lim
x
→
0
f
(
x
)
=
L
1
, and
lim
x
→
0
f
(
x
)
=
L
2
, and prove that
L
1
=
L
2
.]
" (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…
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.