Proof of Limit Law 1 Use the formal definition of a limit to prove that lim ( x , y ) → ( a , b ) ( f ( x , y ) + g ( x , y ) ) = lim ( x , y ) → ( a , b ) f ( x , y ) + lim ( x , y ) → ( a , b ) g ( x , y ) .
Proof of Limit Law 1 Use the formal definition of a limit to prove that lim ( x , y ) → ( a , b ) ( f ( x , y ) + g ( x , y ) ) = lim ( x , y ) → ( a , b ) f ( x , y ) + lim ( x , y ) → ( a , b ) g ( x , y ) .
Solution Summary: The author explains that the function f has the limit L as P(x,y) approaches P_0
Proof of Limit Law 1 Use the formal definition of a limit to prove that
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Assume that
lim f(x, y) = 3
(x,y)→(2,5)
lim g(x, y) = 4
(x,y)-(2,5)
Evaluate the following limit.
lim
(x,y)-(2,5)
ef(x,y)²-6g(x,y)
(Use symbolic notation and fractions where needed.)
limef(x,y)²-6g(x,y) –
=
(x,y)-(2,5)
" (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…
Determine the limit, if it does not exist explain why.
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