Uniqueness of Limits

Definition

Uniqueness of Limits

If a sequence or function has a limit, then that limit is unique. That is, if $lim_{x \to x_{0}} f (x) = L_{1}$ and $lim_{x \to x_{0}} f (x) = L_{2}$ , then $L_{1} = L_{2}$ .

Proof

Uniqueness of Limits

Suppose for the sake of contradiction that a function $f (x)$ has two distinct limits $L_{1}$ and $L_{2}$ as $x \to x_{0}$ , where $L_{1} \neq = L_{2}$ .

Let $ϵ = \frac{∣ L _{1} - L _{2} ∣}{2}$ . Since $L_{1} \neq = L_{2}$ , we have $ϵ > 0$ .

By the definition of a limit, there exists $δ_{1} > 0$ such that for all $x$ : $0 < ∣ x - x_{0} ∣ < δ_{1} ⟹ ∣ f (x) - L_{1} ∣ < ϵ$

Similarly, there exists $δ_{2} > 0$ such that for all $x$ : $0 < ∣ x - x_{0} ∣ < δ_{2} ⟹ ∣ f (x) - L_{2} ∣ < ϵ$

Let $δ = min (δ_{1}, δ_{2})$ . For any $x$ satisfying $0 < ∣ x - x_{0} ∣ < δ$ , both inequalities hold. Using the triangle inequality: $∣ L_{1} - L_{2} ∣ = ∣ L_{1} - f (x) + f (x) - L_{2} ∣ \leq ∣ f (x) - L_{1} ∣ + ∣ f (x) - L_{2} ∣$

Substituting the $ϵ$ bounds: $∣ L_{1} - L_{2} ∣ < ϵ + ϵ = 2 ϵ = ∣ L_{1} - L_{2} ∣$

This results in the contradiction $∣ L_{1} - L_{2} ∣ < ∣ L_{1} - L_{2} ∣$ , proving that $L_{1}$ must equal $L_{2}$ .

Lukas' Notes

Uniqueness of Limits

Definition

Proof

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