Uniqueness of Limit of Function in Metric Space

Proof

Consider a metric space \(X\) with \(a \in X\), in which the function \(f\) is such that \(\lim_{x \to a} f(x) = L_1\) and \(\lim_{x \to a} f(x) = L_2\).

We then assume, by way of contradiction, that \(L_1 \neq L_{2}\), which implies that \(d(L_1, L_2) > 0\) from the properties of the metric. As such, let \(\epsilon = \frac{d(L_1, L_2)}{2}\). From the definition of the limit, there exists a \(\delta_{1}\) such that

and a \(\delta_{2}\) such that

However then for \(\delta = \min\{\delta_1, \delta_2\}\) we have by the triangle inequality that

a contradiction.