Limit of a univariate function

Let $f(x)$ be a function. Suppose we move $x\in \mathbb{R}$ towards $a$ while keeping $x\ne a$. If in this case $f(x)$ approaches a constant value $\alpha$ irrespective of the way how $x$ approaches $a$, we say that $f(x)$ converges to $\alpha$ as $x\to a$ and write $$\underset{x\to a}{lim}f(x)=\alpha$$ or $$f(x)\to \alpha \text{ as }x\to a.$$

Remark. $a$ needs not belong to $\text{dom}(f)$ (the domain of $f$) as long as $x$ can approach $a$ arbitrarily closely. □

But what does this mean exactly? Here's a rigorous definition in terms of what is called the $\epsilon -\delta$ argument.

Definition (Limit of a function)

We say that the function $f(x)$ converges to $\alpha$ as $x\to a$ and write

$$\underset{x\to a}{lim}f(x)=\alpha$$

if the following condition is satisfied.

• For any $\epsilon >0$,  there exists $\delta >0$ such that, for all $x\in \text{dom}(f)$, if $0<|x-a|<\delta$ then $|f(x)-\alpha |<\epsilon$.

In a logical form,

$$\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0,\mathrm{\forall }x\in \text{dom}(f)(0<|x-a|<\delta ⟹|f(x)-\alpha |<\epsilon ).$$

Remark. We are implicitly assuming $\epsilon ,\delta \in \mathbb{R}$. □

Here's how this definition works. Suppose $f(x)\to \alpha$ as $x\to a$. Let us pick any positive real number $\epsilon$. However small this $\epsilon$ may be, if $x$ is sufficiently close to $a$, we can always have $|f(x)-\alpha |<\epsilon$. Here ``sufficiently close'' means that we can pick some sufficiently small positive real number $\delta$ such that $|x-a|<\delta$ implies $|f(x)-\alpha |<\epsilon$. In other words, we move $x$ closer and closer to $a$ until $|f(x)-\alpha |<\epsilon$ holds. Conversely, if this operation is possible for any $\epsilon$, it makes sense to say that $f(x)$ converges to $\alpha$ as $x\to a$.

Example. Consider $f(x)=x^2+1$. We have

$$\underset{x\to 1}{lim}f(x)=2.$$

Let $\epsilon =0.1$. Let us find $\delta >0$ such that $0<|x-1|<\delta$ implies $|f(x)-2|<\epsilon$. Suppose

$$|f(x)-2|<0.1.$$

Then, since $f(x)-2=(x^2+1)-2=x^2-1$,

$$-0.1<x^2-1<0.1$$

so

$$0.9<x^2<1.1.$$

If $x>0$, then

$$\sqrt{0.9}<x<\sqrt{1.1}.$$

Since $\sqrt{0.9}-1=-0.0513\cdots$ and $\sqrt{1.1}-1=0.0488\cdots$, if

$$|x-1|<\sqrt{1.1}-1,$$

then we have

$$|f(x)-2|<0.1=\epsilon .$$

So $\delta$ can be any positive number less than $\sqrt{1.1}-1$. For example, let $\delta =0.04$. Then $|x-1|<0.04$ implies $0.96<x<1.04$, which implies $-0.0784<x^2-1<0.0816$ so

$$|f(x)-2|=|x^2-1|<0.0816<0.1=\epsilon .$$

□

Example. Let us show that $\underset{x\to 1}{lim}2x=2$ by using the $ϵ-\delta$ argument.

Let $\epsilon >0$ be any positive real number. Define $\delta =\frac{\epsilon }{2}$. Then, if $0<|x-1|<\delta$, then we have $|f(x)-2|=|2x-2|=2|x-1|<2\delta =\epsilon$. Therefore $\underset{x\to 1}{lim}f(x)=2$. □