# Can we say a function is "unbounded" when we mean it''s tending to infinity?

I'm watching the Limits series on Khan Academy. In many videos Sal repeatedly says that although some people say that functions that tend to infinity have a limit infinity. (For example, in this video, the says that the function $$y = \frac {2}{x-1}$$ (here's a link to the graph) is unbounded as x approaches 1 from the left side, although "some" people would say that the function is tending towards infinity (i.e., the $$\lim_{x \to c} f(x) = \infty$$.

Over a series of videos, I caught hold of that argument and always said that function don't tend towards infinity; they're just unbounded. But, in a different video (here), Sal essentially says:

$$\lim_{x \to 0} \frac 1{x^2} = \infty$$ $$\lim_{x \to 0^+} \frac 1{x} = \infty$$ $$\lim_{x \to 0^-} \frac 1{x} = -\infty$$

There is an answer to this question that says that unbounded sequences don't always tend to infinity. So it seems that we can't say that the limit of unbounded functions as x tends to some number is infinity -- what I'm confused about, and my question is: When do we ever say that the limit of some function is infinity? Specifically, are the limits of the above functions $$\frac 1x$$ and $$\frac 1{x^2}$$ correct? Or are they unbounded? If we can say that these limits are correct, why can't we say this: $$\lim_{x \to 1^-} \frac 2{x-1} = - \infty$$

• Things that tend to infinity are unbounded, but unbounded things do not always tend to infinity. Tending to infinity means more: it means it is both unbounded and at the same time doesn't "go back down". Commented Nov 30, 2018 at 11:38
• One of the limits is wrong: $\frac 1 {x^{2}}$ is a positive function so it cannot tend to $-\infty$. All other limits are correct. Commented Nov 30, 2018 at 11:46
• @KaviRamaMurthy -- Thank you, edited. Commented Nov 30, 2018 at 12:22
• @Arthur -- Thanks, that explains it. Commented Nov 30, 2018 at 12:23

A necessary condition for a function to have infinite limit is that it is unbounded.

More precisely, if $$\lim_{x\to c}f(x)=\infty$$, then $$f$$ is upper unbounded; if $$\lim_{x\to c}f(x)=-\infty$$, then $$f$$ is lower unbounded. (The limit in the previous statements can also be one-sided.)

One could be more precise and say that $$f$$ must be (upper/lower) unbounded in every punctured neighborhood of $$c$$ (or punctured right/left neighborhood of $$c$$).

This essentially follows from the definition.

However, this condition is by no means sufficient. Consider $$f(x)=\frac{1}{x}\sin\frac{1}{x}$$ Then this function is unbounded in every punctured neighborhood of $$0$$, but its limit is neither $$\infty$$ nor $$-\infty$$.

Restricting to the limit from the right or the left doesn't improve the situation.

Confusing “unbounded” with “has infinite limit” may be very dangerous.

Like Arthur said in the comments, functions being "bounded" and "tending to infinity" mean different things. If $$\lim_{x\to c}f(x)=\infty$$, then it tends to infinity and gets arbitrarily large in the process. Hence it is not bounded, so there does not exist real numbers $$a,b$$ so that $$f(x)\in[a,b]$$ for all $$x$$ in the domain. However, if a function is unbounded it does not necessarily tend to infinity. In fact, the limit need not even exist. Just consider $$f(x)=\sin(1/x)/x$$.