Here is an important lemma.
[MATH]L = \lim_{x \rightarrow a} f(x) \text { and } L < K \implies K \ne \lim_{x \rightarrow a} f(x).[/MATH]
Here is a proof.
[MATH]\text {By definition, for any positive } \epsilon, \ L = \lim_{x \rightarrow a} f(x) \implies[/MATH]
[MATH]\exists \ \delta_1 \text { such that } 0 < |x - a| < \delta_1 \implies 0 \le |f(x) - L| < \epsilon.[/MATH]
[MATH]\therefore 0 < |x - a| < \delta_1 \implies - \epsilon < f(x) - L < \epsilon \implies L - \epsilon < f(x) < L + \epsilon.[/MATH]
[MATH]\text {ASSUME, for purposes of contradiction, that } \lim_{x \rightarrow a} f(x) = K.[/MATH]
[MATH]\therefore \text {By definition, for any positive } \epsilon, \ K = \lim_{x \rightarrow a} f(x) \implies[/MATH]
[MATH]\exists \ \delta_2 \text { such that } 0 < |x - a| < \delta_2 \implies 0 \le |f(x) - K| < \epsilon.[/MATH]
[MATH]\therefore 0 < |x - a| < \delta_2 \implies - \epsilon < f(x) - K < \epsilon \implies K - \epsilon < f(x) < K + \epsilon.[/MATH]
[MATH]\text {Let } \epsilon = \dfrac{K - L}{2} \text { and } \delta = \text {min}(\delta_1,\ \delta_2).[/MATH]
[MATH]\therefore 0 < |x - a| < \delta \implies f(x) < L + \epsilon = L + \dfrac{K - L}{2} = \dfrac{K + L}{2} \implies[/MATH]
[MATH]f(x) < \dfrac{K + L}{2}.[/MATH]
[MATH]\text { And } 0 < |x - a| < \delta \implies K - \epsilon < f(x) \implies K - \dfrac{K - L}{2} < f(x) \implies[/MATH]
[MATH]\dfrac{K + L}{2} < f(x).[/MATH]
[MATH]\therefore 0 < |x - a| < \delta \implies \dfrac{K + L}{2} < f(x) < \dfrac{K + L}{2}, \text { which is impossible.}[/MATH]
[MATH]\text {THUS, } K \ne \lim_{x \rightarrow a} f(x).\ \text {Q.E.D.}[/MATH]
An almost identical proof supports this additional lemma:
[MATH]L = \lim_{x \rightarrow a} f(x) \text { and } L > K \implies K \ne \lim_{x \rightarrow a} f(x).[/MATH]
The two lemmas together prove the theorem
[MATH]L = \lim_{x \rightarrow a} f(x) \text { and } L \ne K \implies K \ne \lim_{x \rightarrow a} f(x).[/MATH]
In English, if a limit exists, it is unique.
