# Do limits evaluated at infinity exist?

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?

• 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 $|f(x)-L| <\epsilon$. – copper.hat Sep 18 '16 at 17:41
• 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. – copper.hat Sep 18 '16 at 17:43
• 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. – Simply Beautiful Art Sep 18 '16 at 22:47

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.

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 $|f(x)-L| < \epsilon$.

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

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.

• 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. – user21820 Sep 19 '16 at 5:10
• @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. – egreg Sep 19 '16 at 8:37
• @egreg: I agree with you completely, but was just pointing out the possibility of such a situation. – user21820 Sep 19 '16 at 8:49