I'm pretty sure there is either something fundamental missing in my understanding of limits, that or I'm completely off mark. Regardless, please help solidify my understanding of limits.

As far as I know, a limit is some value a function, such as f(x), approaches as x gets arbitrarily close to c from either side of the latter.

If this is the case, how can constant functions, such as y=3, have limits? I know the limit of y=3 would be 3 (regardless of what x approaches), but the thing is, y will never approach 5, because it already is 5! You can't get closer to a chair when you are already sitting on it!

Thanks in advance.

  • $\begingroup$ As you yourself wrote in the first part of the question: the limit has to be taken with respect to $x$. So it makes no sense to say "as $y$ approaches". $\endgroup$
    – b00n heT
    Commented Sep 3, 2016 at 22:37
  • 1
    $\begingroup$ The definition of limit formalises a sense of arbitrarily close to but not arbitrarily close to though not equal to $\endgroup$
    – Henry
    Commented Sep 3, 2016 at 22:39
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    $\begingroup$ No variable or function really approaches whatever in analysis: it is only a metaphoric phrase. $\endgroup$
    – Bernard
    Commented Sep 3, 2016 at 22:43
  • $\begingroup$ The definition says it has be get arbitrarily close. It never says at any point it has to be away. $\endgroup$
    – fleablood
    Commented Sep 3, 2016 at 22:53
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    $\begingroup$ Getting closer to the chair doesn't matter. Being arbitrarily close to the chair is what's important. And being in the chair is indeed very very close. $\endgroup$
    – fleablood
    Commented Sep 4, 2016 at 0:01

4 Answers 4


If you do not like the word "approach" let us say the limit predicts where $f(x)$ will end up as $x \to c$.

If $f(x)$ sits there all along on its chair at $5$, surely it makes sense to predict it'll stay there.

  • $\begingroup$ smacks forehead so the limit just predicts where f(c) should be. $\endgroup$
    – user366028
    Commented Sep 3, 2016 at 22:54
  • $\begingroup$ Yes, informally you can think of it in this way. You check $f(x)$ closer and closer to $c$ and based on this you try to infer/predict what $f(c)$ should be. And if $f$ is continuous then the prediction is actually correct. $\endgroup$
    – quid
    Commented Sep 3, 2016 at 22:58
  • $\begingroup$ Very good and smart use of language to convey the right idea about limits +1. $\endgroup$
    – Paramanand Singh
    Commented Sep 4, 2016 at 5:13

The definition says it can get arbitrarily close, but it never says it has to ever be any distance away.

The formal definition: for any $\epsilon > 0$ we can find a $\delta$ so that $|x - x_0| < \delta \implies |f(x) - f(x_0)|< \epsilon$.

This definition certainly holds. Let $\delta = anything$ then $|x - x_0| = anything$, then we have $|f(x) - f(x_0)| = 0 < \epsilon$ for any epsilon.

Yep. That's a limit all right.

  • $\begingroup$ To be honest, I haven't learnt the formal definition of limits yet (I work on that now). But $\endgroup$
    – user366028
    Commented Sep 3, 2016 at 23:08
  • $\begingroup$ I think my problem was that I intuitively thought limits were only to be used to describe the answer of functions when imputed with with a value that would make them undetermined and undefined. Thanks for the clarification. $\endgroup$
    – user366028
    Commented Sep 3, 2016 at 23:19

The definition of limit never actually says you can't hit the value you're approaching. We say that $\lim_{x \to x_0} f(x) = L$ if $$ \forall \epsilon > 0 \ \exists \delta > 0 \ \big[ 0 < |x - x_0| < \delta \implies | f(x) - L | < \epsilon \big] $$

Of course, you could define your own operation, similar to a limit, where the consequent is $0 < | f(x) - L | < \epsilon$ instead. Let's call it a pseudo-limit. It turns out that many of the limit laws that we take for granted fail when you use pseudo-limits.

For example, we expect the sum law to be true (assuming the left side exists): $$\lim_{x \to x_0} f(x) + \lim_{x \to x_0} g(x) = \lim_{x \to x_0} f(x) + g(x)$$

But for pseudo-limits, this can fail. Let $f(x) = \frac{\sin x}{x}$ and $g(x) = 1 - \frac{\sin x}{x}$. We know that as $x \to 0$, we have $f(x) \to 1$ and $g(x) \to 0$. We'd expect the function $(f + g)(x)$ to approach $1$. But $(f + g)(x) = 1$, and so it's a constant function, and has no pseudo-limit.

You can construct similar counterexamples for the other limit laws.


I don't think the confusion here is about limits. I think it's a question about functions. $f(x)$ is not a function. $f: \mathbb R \to \mathbb R$ is a function. What it does, is assigns every real number something in $\mathbb{R}$, uniquely.

For example, your constant function is a function $f: \mathbb{R} \to \mathbb{R}$ such that $f(x)=5$ for all $x \in \mathbb{R}$. In other words, if you give me a real number, say $5$, I will return $3$, since $f(5)=3$.

THE limit of a function does not have any coherence. There are many limits one could take. For example, some limits of the constant function: $\lim_{x \to 0}f(x)=3=\lim_{x \to 1} f(x)=3= \lim_{x \to 2} f(x)=3$ and so on. In other words, the limit here is trivial. Take $\lim_{x \to 0}f(x)$. We want to $f$ to get sufficiently close to $3$. In our case, we need not look far, as for all $x \in \mathbb R$, $f(x)=3$, and so $|f(x)-3|=0$.

As for the last part of your question, limits are indeed unique.

"proof" suppose that $\lim_{x \to y} f(x)=a$, but also that $\lim_{x \to y} f(x)=b$ for $a \neq b$. Then, on what interval can we ensure that $f(x)$ converges? The problem here is that what this limit tells us is that we can get arbitrarily close to different values, but this can't be the case! Take the distance $\frac{a-b}{2}$. Here, even as $x \to y$, $f(x)$ can't get close enough to one of them.

To perhaps aid in intuition, take $g: \mathbb R \to \mathbb R$ defined by $g(x)=\frac{1}{x}$. perhaps you've noticed the asymptote, or the proof that $\lim_{x \to \infty} \frac{1}{x}=0$. Could it be the case that $\lim_{x \to \infty} \frac{1}{x}=-1$? Or any negative number for that matter? Then there would always be the distance between that number and the $x-axis$, a distance that $\frac{1}{x}$ can never satisfy.

I'll leave the $\epsilon-\delta$ to another answer, I just wanted to maybe help untangling this shady business in analysis, without deferring to rigour.


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