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.

• "closer" in this is a loose word, if you're at the point, that's even better :) – Alan Jun 1 '15 at 1:21
• An even more interesting counterexample to the "working definition" is $x\sin\frac1x$ as $x\to0$, which, when $x$ gets closer to $0$, actually reaches $0$, goes away for a bit, comes back to $0$, overshoots, comes back again, and does this infinitely many times. – Rahul Jun 1 '15 at 1:42
• My answer below gives two examples: one in which $f(x)$ gets closer to $7$ as $x$ gets closer to $4$, but in which the limit is $6$, not $7$, and one in which $g(x)$ DOES NOT get closer to $10$ as $x$ gets closer to $4$ and yet $\lim\limits_{x\to4}g(x)=10$. ${}\qquad{}$ – Michael Hardy Jun 1 '15 at 3:22
• Someone commented that he couldn't understand my examples, so I've added a bit of further explanation. – Michael Hardy Jun 1 '15 at 13:37
• Another working definition is: $\lim\limits_{x\to a}f(x)=L$, if when $x\approx a$, we have $f(x)\approx L$. For example, $\lim\limits_{x\to0}\dfrac{\sin x}x=1$, because $\dfrac{\sin(0.0001)}{0.0001}=0.99999999833$. (It's a little vague, but if you want a precise definition you'll want the epsilon-delta.) – Akiva Weinberger Jun 2 '15 at 9:35

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 $x_0$".

• Thank you for answering. I think I am confusing myself by dwelling on 'intuitive' working definitions. I had such a hard time understanding the epsilon delta definition because all i could think about was how it, in some cases, contradicted various working definitions. – Carefullcars Jun 1 '15 at 2:32
• @Carefullcars And that's why the formal delta-epsilon definition is important. It's impossible to make a non-formal/"working" definition that matches it 100% of the time. In fact, we had calculus for centuries before delta-episilon, during which time many people were worried calculus would turn out to be completely wrong. – Ixrec Jun 1 '15 at 7:40
• @Ixrec I don't think people were worried calculus would turn out to be wrong. The thing is that due to lack of rigour people were not even agreeing on what the simplest things were (e.g., what a function is). Nonetheless, the objects of main study were solid enough for it to be clear enough that while a formalism is lacking, it is only a question of time before it is found. People realised the importance of finding a formalism, but not because they needed convincing of consistency. – Ittay Weiss Jun 1 '15 at 7:56
• @Itlay: The objects of study may have been solid enough, but the methods of study were suspect. IIRC, Abel commented how surprising it is that people didn't produce absurd results more often, and essentially attributed it to the fact that people mainly worked with analytic functions, which have much more forgiving properties. – Hurkyl Jun 1 '15 at 15:49
• I've made your intuitive phrasing more precise. =) – user21820 Jun 2 '15 at 8:53

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 $1$.

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 $10$.

• I would say they both are bad examples. I have no idea why you would say it clearly would get closer to 7 since that function doesn't even get larger then 6. And in your second example you say it alternately gets closer and farther away but I don't see it go farther away at all. 10 + 0.001 is just as close to 10 as 10 - 0.001 so I would say it would alternately be closer to 10 and just as close to 10 – Ivo Beckers Jun 1 '15 at 8:56
• @IvoBeckers : Indeed, it does not get larger than $6$, so the limit cannot be $7$. But it does get closer to $7$ as $x$ gets closer to $4$ because it is less than $6$ and getting bigger, so it gets closer to every number that is either $6$ or bigger than $6$. In the second example, as $g(x)$ goes from $10+0.1$ down to $10-0.1$, it gets closer to $10$ as it's going from $10+0.1$ down to $10$ and farther from $10$ as it continuous going downward from $10$ to $10-0.1$. The fact that you have no idea why these functions behave as I said they do shows only that you haven't understood. ${}\ \ {}$ – Michael Hardy Jun 1 '15 at 13:24
• I now understand. I didn't mean to rant or anything. Thanks for clarifying. should I delete my first comment? – Ivo Beckers Jun 1 '15 at 13:33
• @IvoBeckers: You could leave your comment because other people might also have the same question. =) – user21820 Jun 2 '15 at 8:55

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