# Should we write $\lim_\limits{x\to -\infty} \frac{3x^2-1}{x^3+4x+3}=0^-$ instead of just $0$?

This may sounds silly but I have a doubt about a notation. Say I have

$$\lim_{x\to -\infty} \dfrac{3x^2-1}{x^3+4x+3}$$

The limit is obviously zero, for the polynomial at the denominator is of a higher degree than the one at the numerator.

Yet I was thinking: should we (or could we, or must we...) be more precise and write the result as $$0^-$$ instead of just $$0$$, or is that something wrong or nonsensical?

It's like when $$\lim_{x\to 0} \dfrac{1}{x}$$ does not exist unless we specify if it's $$x\to 0^+$$ or $$x\to 0^-$$, but reversed.

• The notation $0^-$ and $0^+$ is reserved for the direction from which we approach the limit. You could use it for "from which side values approach the limit", but it is a not a common approach, plus it doesn't work in your case. Oct 26, 2022 at 16:10
• Your last limit still doesn't exist when you specify right or left side. Oct 26, 2022 at 16:11
• @Randall I meant "not exists" like not the same from left and from right. Then sure it doesn't exist in terms of real numbers, since it's plus or minus infinite. But to me a limit does not exist when it's undefined or indetermned, such as $(-1)^n$ for $n$ to $+\infty$ and so on Oct 26, 2022 at 16:13
• While it's not standard, it's pretty clear what it means: formally, it would mean for every $\epsilon > 0$, there exists $R$ such that whenever $x < R$, $-\epsilon < \frac{3x^2-1}{x^3+4x+3} < 0$. That would then enable you to "compose such limits" -- especially if you started writing for example $\lim_{x\to a^{\ne}} f(x)$ instead of $\lim_{x\to a} f(x)$ for limits at finite points. (But then, since there's no unique choice of what to write on the right hand side, it might make more sense to write something like $f(x) \to 0^-$ as $x \to -\infty$.) Oct 26, 2022 at 17:15
• It's not a standard mathematical notation, but signed zero is a thing in computer floating point. For example, if you open your browser's JavaScript console and enter f = x => (3 * x * x - 1) / (x * x * x + 4 * x + 3); f(-1e150);, you'll probably see -0 as the result.
– Dan
Oct 26, 2022 at 19:20

The limit is a real number, so $$0^-$$ has no meaning in this context. It is usual to preserve the sign when we are in an intermediate step. For instance, we could write something like $$\lim_{x\to -\infty}\dfrac{1}{\dfrac{3x^2-1}{x^3+4x+3}} = \frac{1}{0^-} = -\infty.$$