1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Sep 3 '16 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 Sep 3 '16 at 22:39
  • 1
    $\begingroup$ No variable or function really approaches whatever in analysis: it is only a metaphoric phrase. $\endgroup$ – Bernard Sep 3 '16 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 Sep 3 '16 at 22:53
  • 1
    $\begingroup$ 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 Sep 4 '16 at 0:01
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ smacks forehead so the limit just predicts where f(c) should be. $\endgroup$ – user366028 Sep 3 '16 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 Sep 3 '16 at 22:58
  • $\begingroup$ Very good and smart use of language to convey the right idea about limits +1. $\endgroup$ – Paramanand Singh Sep 4 '16 at 5:13
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ To be honest, I haven't learnt the formal definition of limits yet (I work on that now). But $\endgroup$ – user366028 Sep 3 '16 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 Sep 3 '16 at 23:19
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.