Why does the definition of limits of a function have strict inequality?

Solution 1:

Excellent question. In fact, it turns out that the definition with $\leq$ is equivalent.

Suppose that $f$ approaches the limit $l$ near $a$ using the standard definition (the one with $<$), so that for every $\epsilon >0$ there is some $\delta > 0$ such that, for all $x$, if $0<|x-a|<\delta$, then $|f(x)-l|<\epsilon$.

Then it is also the case that for any $\epsilon>0$, there is some $\gamma>0$ such that, for all $x$, if $0<|x-a|\leq\gamma$, then $|f(x)-l|\leq\epsilon$. What is the $\gamma$ we can produce? Well, taking the $\delta$ we know exists for this $\epsilon$ from the previous paragraph, we can choose $\gamma=\frac{\delta}{2}$, because $$0<|x-a|\leq\frac{\delta}{2}\implies 0<|x-a|<\delta,$$ and we know that any $x$ with $0<|x-a|<\delta$ satisfies $|f(x)-l|<\epsilon$, so certainly any such $x$ will also satisfy $|f(x)-l|\leq\epsilon$.

(In case it's not clear, the fact that we've used the letter $\gamma$ here is of no consequence; I only did so to help keep the two situations mentally separate.)

Thus, if $f$ approaches the limit $l$ near $a$ using the $<$ definition, it will also approach the limit $l$ near $a$ using the $\leq$ definition.

Showing the other implication is similar.

Solution 2:

Let's look at the definitions more carefully. Suppose $f:\mathbb{R}\to \mathbb{R}$ and at point $a$ the "limit" is $L$.

1. $$(\forall \epsilon>0)(\exists \delta>0)(\forall x\in \mathbb{R})\ (0<\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon)$$
2. $$(\forall \epsilon>0)(\exists \delta>0)(\forall x\in \mathbb{R})\ (0<\left|x-a\right|\le \delta\implies \left|f(x)-L\right|\le\epsilon)$$

The two are equivalent. Indeed I will prove $1\implies 2$ and $2\implies 1$

Suppose 1. Let $\epsilon>0$. Then $\exists \delta>0$ so that $\forall x\in \mathbb{R}$, $$\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon$$ Choose $0<\delta^{\prime}<\delta$. Then, $$0<\left|x-a\right|\le \delta^{\prime}\implies 0<\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon\implies \left|f(x)-L\right|\le\epsilon$$

Suppose 2. Let $\epsilon>0$ and choose $0<\epsilon^{\prime}<\epsilon$. Then $\exists \delta>0$ so that $\forall x\in \mathbb{R}$, $$\left|x-a\right|\le \delta\implies \left|f(x)-L\right|\le \epsilon^{\prime}$$ Then, $$0<\left|x-a\right|<\delta\implies \left|x-a\right|\le\delta\implies \left|f(x)-L\right|\le \epsilon^{\prime}\implies \left|f(x)-L\right|<\epsilon$$

Note: As Zev point above we can always choose $\delta^{\prime}=\frac{\delta}2$ and $\epsilon^{\prime}=\frac{\epsilon}2$. You may also want to ask why there are only implications and not equivalencies in the 2 definitions

Solution 3:

The two ways of writing the definition you have written above are equivalent.

Suppose for every $\epsilon>0$, there is a $\delta>0$ such that $0<|x-a|<\delta$ implies $|f(x)-l|<\epsilon$. Then $0<|x-a|\leq\frac{\delta}{2}$ implies $|f(x)-l|\leq\epsilon$, with $\frac{\delta}{2}>0$.

Conversely, suppose for every $\epsilon>0$, there is a $\delta>0$ such that $0<|x-a|\leq\delta$ implies $|f(x)-l|\leq\epsilon$. Then for every $\epsilon>0$, there is a $\delta>0$ such that $0<|x-a|\leq\delta$ implies $|f(x)-l|\leq\frac{\epsilon}{2}$, since $\frac{\epsilon}{2}>0$; hence $0<|x-a|<\delta$ implies $|f(x)-l|<\epsilon$.