Extension: Uniqueness of One-Sided Limits

Limits, when they exist, are unique. (yay!)

Consider $\lim_{x\to c}f(x)$, where $c$ may be $\pm\infty$ as well. Suppose there were two limits, $L_1$ and $L_2$.

Intuitively, there can only be one limit - if $f(x)$ is getting arbitrarily close to, say, $L_1$, then it cannot be getting arbitrarily close to $L_2$ as well! Using this intuition, let's formalize it.

Recall the definition of $\lim_{x\to c}f(x)=L$ (the case for $\pm\infty$ is similar):

For every $\varepsilon>0$, there exists a $\delta>0$ such that $0<|x-c|<\delta\Rightarrow|f(x)-L|<\varepsilon$.

Informally, this means that you give me an $\varepsilon>0$ and challenge me to show that $f(x)$ gets within $\varepsilon$ of $L$. I respond by giving you the $\delta>0$, claiming that if $x$ is within $\delta$ of $c$, then $f(x)$ is within $\varepsilon$ of $L$. The limit existing means that I can find a $\delta>0$ for every $\epsilon>0$ that you give me.

Now on to the proof:

We want to use the definition of a limit to formalize our intuition. That is, we want to show that if $f$ were somehow approaching two limits $L_1\ne L_2$, then we'd get a contradiction somewhere.

So let's suppose that $\lim_{x\to c}f(x)=L_1$ and $\lim_{x\to c}f(x)=L_2$. Then using the definition of limit, for any $\varepsilon>0$, there exists $\delta_1,\delta_2>0$ such that

$0<|x-c|<\delta_1\Rightarrow|f(x)-L_1|<\varepsilon$ and $0<|x-c|<\delta_2\Rightarrow|f(x)-L_2|<\varepsilon$.

Now take $\delta=\min\{\delta_1,\delta_2\}$. Then $$0<|x-c|<\delta\Rightarrow|f(x)-L_1|<\varepsilon\text{ and }|f(x)-L_2|<\varepsilon.$$ Now we can use the triangle inequality, to write $$2\varepsilon>|L_1-f(x)|+|f(x)-L_2|\ge|L_1-L_2|,$$ so that $$|L_1-L_2|<2\varepsilon.$$ But this can't work for every $\varepsilon>0$ ! (eg. take $\varepsilon=\frac{|L_1-L_2|}{2}$.) A contradiction, so we must have $L_1=L_2$.