AP Calculus AB and BC

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

Q: What function is |x|/x?

It is the sign function on x \neq 0: 1 for positive inputs, -1 for negative ones, undefined at 0. The graph is two horizontal rays with a gap between them, which is the cleanest picture of a jump discontinuity there is.

Q: Why does the two-sided limit not exist?

The right side gives 1 and the left side gives -1. A two-sided limit exists only when both one-sided limits exist and agree, so a mismatch of 1 against -1 ends the question. Note that both one-sided limits exist perfectly well.

Q: Does this show up in derivative questions?

Yes. The difference quotient for f(x) = |x| at 0 is \frac{|h|}{h}, whose one-sided limits are 1 and -1. That mismatch is the proof that |x| has no derivative at 0, and it is why the graph shows a corner there.

The limit of the absolute value of x over x as x approaches 0 from the right is 1. For positive x the absolute value leaves x alone, so the quotient equals 1 at every input on that side. From the left it equals negative 1, which makes 0 a jump discontinuity.

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

Settled by evaluating the piecewise definition.

Splitting the absolute value into cases

Absolute value is a piecewise rule, not an operation you push through the algebra. It reads $|x| = x$ when $x \geq 0$ and $|x| = -x$ when $x < 0$ . A limit from the right uses only inputs with $x > 0$ , so only the first case is ever in play.

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

The function is not close to 1 on that side, it is equal to 1 at every one of those inputs. A function that is constant on the side being approached has that constant as its one-sided limit.

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

That $x = 0$ itself is outside the domain changes nothing. A limit describes the approach and never the value at the point, which is precisely why limits can detect a hole or a jump that substitution cannot see.

Why substitution fails, and why 0 over 0 misleads here

Substituting 0 gives $\frac{0}{0}$ , the form that normally starts a search for a common factor or a pair of derivatives. Both of those moves waste time on this problem.

Factoring has nothing to grip, since $|x|$ is not a polynomial. L'Hopital's rule is worse: the derivative of $|x|$ is $\frac{x}{|x|}$ for $x \neq 0$ , so the rule hands back $\frac{x}{|x|}$ , the reciprocal of what you started with and no simpler.

0 over 0 says look closer, not apply a rule

The form only reports that substitution was inconclusive. What resolves it comes from what the expression is built from. Polynomials point to factoring, roots point to the conjugate, and absolute values or piecewise rules point to case work.

Once the correct case is selected, nothing indeterminate is left. On $x > 0$ the expression is the constant 1, and reading a constant off is a substitution that now works rather than a limit technique.

The mistake students make

Cancelling the bars and claiming $\frac{|x|}{x} = 1$ for all $x$ . That identity holds only on the right of 0. On the left the quotient is $-1$ .
Reporting that the limit does not exist. This one-sided limit is a clean 1. The two-sided limit is the one that fails, since the sides give 1 and $-1$ .
Reaching for L'Hopital's rule on sight of $\frac{0}{0}$ without checking whether differentiating gets anywhere.
Calling $x = 0$ a removable discontinuity. The gap is a jump of height 2, and no value assigned at 0 can close it.

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

What function is |x|/x?

It is the sign function on $x \neq 0$ : 1 for positive inputs, $-1$ for negative ones, undefined at 0. The graph is two horizontal rays with a gap between them, which is the cleanest picture of a jump discontinuity there is.

Why does the two-sided limit not exist?

The right side gives 1 and the left side gives $-1$ . A two-sided limit exists only when both one-sided limits exist and agree, so a mismatch of 1 against $-1$ ends the question. Note that both one-sided limits exist perfectly well.

Does this show up in derivative questions?

Yes. The difference quotient for $f(x) = |x|$ at 0 is $\frac{|h|}{h}$ , whose one-sided limits are 1 and $-1$ . That mismatch is the proof that $|x|$ has no derivative at 0, and it is why the graph shows a corner there.