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Proof

(related to Proposition: Limit of a Function is Unique If It Exists)

Context

• Let $D\subseteq\mathbb R$ be a subset of real numbers.
• Let $f:D\to\mathbb R$ be a function.
• Let $a\in\mathbb D.$

Hypothesis

• Assume, the limit $\lim_{x\to a} f(x)$ exists, but it not unique, e.g. $\lim_{x\to a} f(x)=L$ and $\lim_{x\to a} f(x)=M$ but $L\neq M.$

Implications

• Since $L\neq M$, then $|M - L| > 0.$
• By the definition of limit, there is a $\delta_1 > 0$ such that for all $x\in D$ satisfying $0 < |x - a| < \delta_1,$ it follows that $|f(x) - L| < \frac{|M-L|}{2}.$
• Similarly, there is a $\delta_2 > 0$ such that for all $x\in D$ satisfying $0 < |x - a| < \delta_2,$ it follows that $|f(x) - M| < \frac{|M-L|}{2}.$
• Take the minimum $\delta:=\min(\delta_1,\delta_2)$ and select $x\in D$ with $0 < |x-a| < \delta.$
• Then it follows by the triangle inequality that $$\begin{array}{rcl}|M-L|&=&\frac{|M-L|}{2}+\frac{|M-L|}{2}\\ &>& |f(x) - L| + |f(x) - M|\\ &=&|f(x)-L| + |M-f(x)|\\ &\ge&|f(x)-L+M-f(x)|\\ &=&|M-L|.\end{array}$$
• This leads to the contradiction $|M-L| > |M-L|.$

Conclusion

• Therefore, $L\neq M$ must be wrong and we have $L=M.$
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