09 - Uniqueness of limits

Uniqueness of limits

Let $(a_{n})$ be a convergent sequence then the limit is unique

Proof

Assume that $a_{n} \to L_{1}$ and $a_{n} \to L_{2}$ as $n \to \infty$
Suppose for a contradiction that $L_{1} \neq = L_{2}$
Let $ϵ = \frac{∣ L _{1} - L _{2} ∣}{2} > 0$
Since $a_{n} \to L_{1}$ as $n \to \infty$ , there $N_{1}$ such that
$if n \geq N_{1} then ∣ a_{n} - L_{1} ∣ < ϵ$
Also, since $a_{n} \to L_{2}$ as $n \to \infty$ , there is $N_{2}$ such that
$if n \geq N_{2} then ∣ a_{n} - L_{2} ∣ < ϵ$
For $n \geq max {N_{1}, N_{2}}$ we have $∣ a_{n} - L_{1} ∣ < ϵ$ and $∣ a_{n} - L_{2} ∣ < ϵ$
$∣ L_{1} - L_{2} ∣ = ∣(L_{1} - a_{n}) + (a_{n} - L_{2})∣ \leq ∣ L_{1} - a_{n} ∣ + ∣ a_{n} - L_{2} ∣ (by the triangle inequality) < 2 ϵ = ∣ L_{1} - L_{2} ∣$
This is a contradiction hence $L_{1} = L_{2}$

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09 - Uniqueness of limits

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