6
$\begingroup$

Here is some limit: $$\lim_{x \to b} f(x)$$

We know that for a limit to exist, we must have $$\lim_{x \to b+} f(x) = \lim_{x \to b-} f(x)$$

So I am confused because, when $b=+\infty$ we can only evaluate this limit from the left side and not the right side. We can't approach infinity from a higher infinity. Does this mean that limits evaluated at infinity don't exist and therefore none of the limit laws like addition apply to limits evaluated at infinity?

EDIT: So does that mean I can use the limit laws such as addition, composition, etc on limits evaluated at infinity as long as the limits tend to a finite value?

I.e.

$$\lim_{x \to \infty} [f(x) + g(x)] = \lim_{x \to \infty} f(x) + \lim_{x \to \infty} g(x)$$ as long as both of the separate limits are some finite value?

And so on, for multiplication, composition, etc?

$\endgroup$
  • 4
    $\begingroup$ Generally, the limit as $x \to \infty$ is defined slightly differently in that we say that $\lim_{x \to \infty} f(x) =L$ iff for all $\epsilon>0$ there is some $N$ such that if $x \ge N$ then $|f(x)-L| <\epsilon$. $\endgroup$ – copper.hat Sep 18 '16 at 17:41
  • $\begingroup$ Defining the limit in terms of right & left limits is a little restrictive, it is better to use an '$\epsilon$' criterion as in the previous comment. $\endgroup$ – copper.hat Sep 18 '16 at 17:43
  • $\begingroup$ What if you could find a number greater than $+\infty$...? What then? As far as I see it, it wouldn't make any difference on the limit. $\endgroup$ – Simply Beautiful Art Sep 18 '16 at 22:47
6
$\begingroup$

The notion of limit from the left/right side simply makes no sense if $x\rightarrow\infty$ or $x\rightarrow -\infty$. The rule you're refering to is only valid when $x$ tends to a (finite) number $b$.

However limits at infinity do exist. Their definition is different from limits at $b\in\mathbb{R}$, but the same algebraic rules apply to them.

$\endgroup$
7
$\begingroup$

Limits at infinity are defined in a different way in standard analysis.

Concretely, we will say that $\lim_{x\to \infty} f(x) = L$ if for every $\epsilon >0$ there is a $M\in\mathbb{R}$ such that for every $x>M$ we have that $|f(x)-L| < \epsilon$.

$\endgroup$
  • 1
    $\begingroup$ \inf should be \infty but this stupid site won't let me do an edit with fewer than 6 characters changed. $\endgroup$ – R.. Sep 19 '16 at 4:37
  • $\begingroup$ @R..: I'm glad I'm not the only one... $\endgroup$ – Mehrdad Sep 19 '16 at 4:55
  • $\begingroup$ @R..: When that happens, make some other minor edits like adding spaces. =) $\endgroup$ – user21820 Sep 19 '16 at 5:09
4
$\begingroup$

Left and right limits are special cases of the general limit concept, and only make sense for limits $x\to b$ when $b\in{\mathbb R}$. We need them in cases when a function is defined only for $x<b$, or if $f$ is given by different expressions for $x<b$ and $x>b$. The statement $$\lim_{x\to b} f(x)=\alpha\quad\Leftrightarrow\quad \lim_{x\to b-}=\alpha\quad\wedge\quad \lim_{x\to b+}=\alpha\ $$ is a proposition, and not a definition.

$\endgroup$
  • 1
    $\begingroup$ This is not necessarily true. Some introductory texts or courses on real analysis might well define the two-sided limit as the conjunction of the one-side limits. $\endgroup$ – user21820 Sep 19 '16 at 5:10
  • $\begingroup$ @user21820 In my opinion it's pedagogically very wrong to define a two-sided limit as the conjunction of one-sided limits. One can compute a two-sided limit using both one-sided limits, which is quite a different thing. $\endgroup$ – egreg Sep 19 '16 at 8:37
  • $\begingroup$ @egreg: I agree with you completely, but was just pointing out the possibility of such a situation. $\endgroup$ – user21820 Sep 19 '16 at 8:49

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.