Wrong intuitive understanding of a limit?

In my textbook, before introducing the epsilon delta definition, they gave a working definition of what a limit is. The definition sounded something like this "$\lim \limits_{x \to a}f(x) = L$, if when $x$ gets closer to $a$, $f(x)$ gets closer to $L$"

But is that always the case with limits? What if $f(x) = 4,$ then we have $\lim \limits_{x \to 2}f(x) = 4$, but it is never true that when x gets closer to 2, f(x) gets closer to 4. Maybe instead we should say: "$\lim \limits_{x \to a}f(x) = L$, if when $x$ gets closer to $a$, $ f(x)$ gets closer to or equals $L$".

Please correct me if I'm wrong. I'm pretty new to this stuff. Btw, i understand that the epsilon delta definition has the constant function limit case covered, but I'm more interested in the working definition.

Solution 1:

You are correct. You present a nuance of the 'working definition', a special case where "getting closer" is misleading. You should interpret "getting closer" as "getting as close as you like". This is a more accurate 'working definition' in any case. Then the constant function scenario works just fine. You can get as close as you like to the constant, quite trivially.

The "getting closer" definition makes it sounds as if a limit is somehow an indefinite thing, moving around, getting closer to things. This is misleading. A better intuition is "the limit at $x_0$ of $f$ is $L$" is "the value $f(x)$ can be made as close as you like to $L$ for all $x$ that is sufficiently close to but not equal to $x_0$".

Solution 2:

Probably a better intuitive definition is $f(x)$ can be made arbitrarily close to $L$ by making $x$ close enough to $a$.

You avoid the awkwardness about the constant case. Additionally this emphasizes that it is for ALL $\epsilon$. It's not just that $f(x)$ gets "closer", it's that it can be made as close as you'd like.

For a counterexample of "gets closer" is $\lim\limits_{x\to 0} -x^2 $. I propose that $\lim\limits_{x\to 0} -x^2 =1$ because as $x$ gets closer to $0$ then $-x^2$ gets closer to $1$. However it doesn't get within $\frac12$ of $1$.

Solution 3:

when $x$ gets closer to $a$, $f(x)$ gets closer to $L$

That is wrong. Consider two examples:

• Let $f(x) = 6-(x-4)^2$. Clearly $f(x)$ never gets bigger than $6$, so the limit cannot be $7$, but $f(x)$ gets closer to $7$ (and to all numbers that are bigger than $6$) as $x$ gets closer to $4$, but $\lim\limits_{x\to4}f(x)$ is $6$, not $7$.

• Suppose that as $x$ approaches $4$, $g(x)$, depending continuously on $x$, goes up to $10+0.1$, then down to $10-0.1$, then up to $10+0.01$, then down to $10-0.01$, then up to $10+0.001$, then down to $10-0.001$, etc. Then $\lim\limits_{x\to4} g(x)=10$. But $g(x)$ does not keep getting closer to $10$, but gets alternately closer and farther away: As it's going down from $10+0.1$ to $10$ it's getting closer to $10$, and as it continues going downward from $10$ to $10-0.1$, it's getting farther from $10$.

Solution 4:

The 'working definition' is actually NOT working and you shoud NOT use it. To make it work replace it with

$f$ gets arbitrarily close to $L$ if $x$ sufficiently approaches $a$

where 'gets arbitrarily close' does not mean just $f$ may get closer to $L$ but rather that it certainly will not get apart from $L$ farther than any arbitrarily set distance.

In other words

$f$ gets arbitrarily close to $L$ and stays there