Introduction
Have you ever written down a vague answer to a limit problem like this?
$$\lim_{x \to -1^-} \frac{1}{x+1} = ?$$"The denominator goes to 0, so isn't it ∞?" — if you write \(+\infty\), you are wrong. The correct answer is \(-\infty\).
The key here is sign. Even when the denominator goes to 0, the result can be \(+\infty\) or \(-\infty\) depending on which side it approaches 0 from. In this post we start with what a symbol like \(x \to c^-\) means, and then track the signs of the numerator and denominator one term at a time so that you can find the answer without any room for error.
Start with the meaning of the symbols
First, let's nail down exactly what the superscripts ⁻ and ⁺ mean.
- \(x \to c^-\) : \(x\) approaches \(c\) from the side smaller than \(c\) (approaching from the left)
- \(x \to c^+\) : \(x\) approaches \(c\) from the side larger than \(c\) (approaching from the right)
Why does this "direction" matter? In a fraction, when the denominator goes to a very small number the fraction blows up, but dividing by a small positive number gives a large positive value (\(+\infty\)), while dividing by a small negative number gives a large negative value (\(-\infty\)). Even for the same "going to 0," the sign decides the result.
You only need to carry one tool: set \(x = c \pm \varepsilon\). Here \(\varepsilon\) (epsilon) is a very small positive number. For a left approach put \(x = c - \varepsilon\), for a right approach put \(x = c + \varepsilon\), and just track the sign.
Example 1: \(\lim\limits_{x \to -1} \dfrac{1}{x+1}\)
Let's check what sign the denominator \((x+1)\) has from each direction. Here \(c = -1\).
Left approach \((x \to -1^-)\)
\(x\) is a number slightly smaller than −1. Let me set \(x = -1 - \varepsilon\). (\(\varepsilon > 0\))
$$ x + 1 = (-1 - \varepsilon) + 1 = -\varepsilon \quad [\text{negative}] $$The denominator is a very small negative number. What happens when you divide 1 by a small negative number?
$$ \frac{1}{x+1} = \frac{1}{-\varepsilon} \quad\longrightarrow\quad -\infty $$If it doesn't click, check with concrete numbers. If \(x = -1.001\) then \(x+1 = -0.001\) and \(\dfrac{1}{-0.001} = -1000\). As \(\varepsilon\) gets smaller, this drops endlessly to \(-10^4, -10^5, \dots\) → \(-\infty\)
Right approach \((x \to -1^+)\)
This time it's a number slightly larger than −1, \(x = -1 + \varepsilon\).
$$ x + 1 = (-1 + \varepsilon) + 1 = +\varepsilon \quad [\text{positive}] $$The denominator is a very small positive number, so
$$ \frac{1}{x+1} = \frac{1}{+\varepsilon} \quad\longrightarrow\quad +\infty $$Check: if \(x = -0.999\) then \(x+1 = 0.001\), and \(\dfrac{1}{0.001} = +1000\). → \(+\infty\)
Conclusion
From the left it is \(-\infty\), from the right it is \(+\infty\) — the left-hand limit and the right-hand limit have different signs.
$$ \lim_{x \to -1^-}\frac{1}{x+1} = -\infty, \qquad \lim_{x \to -1^+}\frac{1}{x+1} = +\infty $$Since the two differ, the limit without a direction, \(\displaystyle\lim_{x \to -1}\dfrac{1}{x+1}\), does not exist. That's why writing a vague \(+\infty\) at the start was wrong.
Example 2: Track both the numerator and the denominator
When the numerator also contains the variable, you check the sign of the numerator and the denominator separately and then combine them.
$$ \lim_{x \to 2^-} \frac{x-3}{x-2} $$Since this is a left approach, set \(x = 2 - \varepsilon\). (\(\varepsilon > 0\))
Numerator \(x - 3\):
$$ x - 3 = (2 - \varepsilon) - 3 = -1 - \varepsilon \quad [\text{negative}] $$(For \(x\) near 2, the numerator is around \(2-3=-1\), so it is negative.)
Denominator \(x - 2\):
$$ x - 2 = (2 - \varepsilon) - 2 = -\varepsilon \quad [\text{a very small negative number}] $$Now just combine the signs. It's (negative) ÷ (small negative), so
$$ \frac{x-3}{x-2} = \frac{(\text{negative})}{(\text{small negative})} = (\text{positive}) \quad\longrightarrow\quad +\infty $$The numerator stays finite near \(-1\) while only the denominator goes to 0, so the magnitude grows without bound, and negative ÷ negative makes the sign positive → \(+\infty\). When you track the sign term by term, there is no room for confusion.
Try it yourself
Use the c slider to move the position of the asymptote, and watch in both the graph and the analysis panel how the sign of the denominator \((x-c)\) splits between the left approach and the right approach.
Even as you move c, the pattern is the same — for the form \(f(x) = \dfrac{1}{x-c}\), the left side is \(-\infty\) and the right side is \(+\infty\). If the sign of the numerator changes or the degree of the denominator differs, this pattern changes too, so it's always safe to make a habit of checking directly with ε.
Key takeaways
Procedure for tracking the sign of left-hand and right-hand limits
- Match the direction and set \(x = c - \varepsilon\) (left) or \(x = c + \varepsilon\) (right). (\(\varepsilon\) is a small positive number.)
- Check the sign of each of the numerator and the denominator separately.
- Combine the signs — (neg)/(neg)=pos, (pos)/(neg)=neg, (neg)/(pos)=neg.
- If the denominator goes to \(0^-\) (small negative) it diverges toward \(-\infty\); if it goes to \(0^+\) (small positive) it diverges toward \(+\infty\).
- If the left-hand limit ≠ the right-hand limit, then the limit without a direction does not exist.
| Side the denominator goes to | Numerator sign | Result |
|---|---|---|
| \(0^+\) | + | \(+\infty\) |
| \(0^+\) | − | \(-\infty\) |
| \(0^-\) | + | \(-\infty\) |
| \(0^-\) | − | \(+\infty\) |
Studying this with AI
To get the hang of limit signs, you have to see the process of "why that sign comes out." If you ask an AI for the process — something like "track the sign of (x−3)/(x−2) at x→2⁻ with ε and explain it" — you get a step-by-step trace instead of just an answer. That said, AI sometimes gets the sign or direction confused, so it's best to cross-check by looking at the graph directly with the interactive above.
Wrapping up
The root of sign mistakes in limits is the habit of vaguely writing "∞".
- \(x \to c^-\) and \(x \to c^+\) mean the direction of approach.
- Set \(x = c \pm \varepsilon\) and check the numerator's and denominator's signs term by term.
- If the left-hand and right-hand limits differ, the limit without a direction does not exist.
Once you make this procedure a habit, sign mistakes drop sharply.
Related reading — Why Doesn't √(A−B) Equal √A−√B? · What Are Trigonometric Functions Even For? · Why Does the Arctangent Show Up? · Computation check: Wolfram Alpha