AP Calculus AB and BC

Limit of |x|/x as x Approaches 0 from the Left

Q: Is the function continuous anywhere?

Everywhere except 0. On x 0 it is the constant 1, and constants are continuous. Only x = 0 breaks, and it breaks as a jump rather than as a hole, so no assigned value repairs it.

Q: How should I write this up for a graph question?

State both sides and then the conclusion: \lim_{x \to 0^-} \frac{|x|}{x} = -1 and \lim_{x \to 0^+} \frac{|x|}{x} = 1, so \lim_{x \to 0} \frac{|x|}{x} does not exist. Rubrics look for the mismatch as the reason, not only for the conclusion.

The limit of the absolute value of x over x as x approaches 0 from the left is negative 1. For negative x the absolute value returns negative x, which is a positive number, so the quotient is negative 1 at every input on that side. The right side gives 1, so the two-sided limit does not exist.

\lim_{x \to 0^-} \frac{|x|}{x} = -1

Settled by evaluating the piecewise definition.

Using the negative branch of the definition

For $x < 0$ the definition says $|x| = -x$ . That expression looks negative and is not: negating a negative input returns a positive number, which is the entire purpose of the rule. At $x = -0.4$ it gives $|-0.4| = 0.4$ .

\frac{|x|}{x} = \frac{-x}{x} = -1 \quad \text{for } x < 0

So on the whole interval to the left of 0 the function is the constant $-1$ . One value is worth checking by hand: at $x = -0.001$ the quotient is $\frac{0.001}{-0.001}$ , which is $-1$ , and the same arithmetic repeats at every input closer in.

\lim_{x \to 0^-} \frac{|x|}{x} = -1

Sign is the only moving part. The numerator stays positive on both sides of 0 because an absolute value is never negative, while the denominator carries the sign of $x$ , so the quotient flips exactly where $x$ does.

Why substitution fails, and what the 0 over 0 form hides

Substitution gives $\frac{0}{0}$ . That form is indeterminate in the strict sense of settling nothing, and this example shows how little it settles: the same expression at the same point produces $\frac{0}{0}$ from both sides, with limit $-1$ from one and $1$ from the other.

No algebra resolves it, because there is no single formula waiting to be simplified. The absolute value is two formulas sharing one symbol, and picking the branch that matches the side of the approach is the whole method.

The side is data, not decoration

In most limit problems the superscript is a detail. In any problem holding an absolute value, a piecewise rule, or a square root of a square, it selects which formula you are allowed to use, so answering without it answers a different question.

Topic 1.10 asks students to classify discontinuities, and one-sided limits that both exist and disagree are the definition of the jump case. Here the jump has height 2, from $-1$ up to $1$ .

The mistake students make

Reading $-x$ as a negative number and concluding the numerator is negative. With $x$ negative, $-x$ is positive, and $|x|$ is never negative for any input.
Writing 1 out of symmetry with the right-hand answer. The numerator is unaffected by the side while the denominator changes sign, so the answer flips.
Cancelling to 1 before checking the side. The identity $\frac{|x|}{x} = 1$ holds only for $x > 0$ .
Concluding that because the two-sided limit fails, the one-sided limits fail too. Both are finite and both exist. They simply disagree.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Why is |x| equal to -x on the left?

Because absolute value reports distance from 0, which is never negative. When $x$ is already negative, the way to report a positive distance is to negate it, so $|-3| = -(-3) = 3$ . The minus sign is an instruction, not a claim about the sign of the result.

Is the function continuous anywhere?

Everywhere except 0. On $x < 0$ it is the constant $-1$ and on $x > 0$ it is the constant 1, and constants are continuous. Only $x = 0$ breaks, and it breaks as a jump rather than as a hole, so no assigned value repairs it.

How should I write this up for a graph question?

State both sides and then the conclusion: $\lim_{x \to 0^-} \frac{|x|}{x} = -1$ and $\lim_{x \to 0^+} \frac{|x|}{x} = 1$ , so $\lim_{x \to 0} \frac{|x|}{x}$ does not exist. Rubrics look for the mismatch as the reason, not only for the conclusion.