The Absolute-Value Difference Quotient at Its Corner

The Difference Quotient at Zero

The function \(f(x)=|x|\) is defined at zero, and its value there is \(f(0)=0\). To ask whether it has a derivative at zero, we examine how the output changes relative to a small change \(h\) in the input. The crucial point is that the absolute-value expression behaves differently according to whether \(h\) is positive or negative.

The derivative at zero, if it exists, is the limit of the difference quotient as the nonzero increment \(h\) approaches zero:

For this function, \(f(0+h)=|h|\) and \(f(0)=0\). Therefore the quotient to evaluate is

The condition \(h\ne 0\) matters: the difference quotient itself has \(h\) in its denominator, so it is not defined at \(h=0\). The derivative question concerns the limit as nonzero values of \(h\) approach zero.

Separate Positive and Negative Increments

The absolute value of a real number is its nonnegative distance from zero. In particular, \(|h|=h\) when \(h>0\), and \(|h|=-h\) when \(h<0\). Substituting each expression into the quotient gives two different results.

Thus every positive increment, however small, gives quotient \(1\); every negative increment, however small, gives quotient \(-1\). These are not merely values the quotient approaches: on each side, it is exactly constant.

The one-sided difference quotient limits are therefore

The first limit describes increments approaching zero from above, and the second describes increments approaching from below. To have a two-sided limit, the quotient must approach the same real number from both sides. Here the two values differ, so the derivative limit does not exist.

Worked Example: Simplifying the Quotient on Each Side

Why the Unequal Limits Prove Nondifferentiability

It is useful to make the final inference precise. A limit at zero, if it exists, must be approached by the quotient for both positive and negative increments. Suppose, for contradiction, that the limit exists and equals some real number \(L\). By the definition of a finite limit, for \(\varepsilon=\frac12\) there must be a \(\delta>0\) such that

Choose a positive \(h\) with \(0<h<\delta\), for example \(h=\frac{\delta}{2}\). The quotient is \(1\), so the assumed limit condition gives

Now choose \(h=-\frac{\delta}{2}\), which is negative and also satisfies \(0<|h|<\delta\). The quotient is \(-1\), so the same limit condition gives

But the triangle inequality then implies

This would say \(2<1\), which is impossible. Therefore no such \(L\) exists. The difference quotient has no two-sided limit at zero, and, by the definition of the derivative, \(f(x)=|x|\) is not differentiable at zero.

Worked Example: Checking the Quotient Along Sequences

Worked Example: Comparing the Secant Slopes from Either Side

A Common Sign Mistake

A frequent error is to use \(|h|=h\) for negative \(h\) as well as positive \(h\). That substitution would produce

The error is in the first equality: when \(h<0\), the absolute value is \(-h\), not \(h\). For example, if \(h=-2\), then \(|h|=2\), whereas \(h=-2\). The correct calculation is

A reliable check is to substitute a particular negative increment before simplifying. With \(h=-2\), the original quotient is \(\frac{|-2|}{-2}=\frac{2}{-2}=-1\), not \(1\). Keeping the signed denominator is essential.

What the Calculation Establishes

The entire conclusion follows from the difference quotient and the definition of absolute value. For positive increments the quotient is exactly \(1\); for negative increments it is exactly \(-1\). Consequently, the one-sided limits are different, the two-sided limit does not exist, and \(f'(0)\) is undefined.

This calculation is a useful model for derivative questions at points where a formula changes behavior. The immediate task is not to guess a slope from a sketch, but to form the quotient, respect the sign of the increment, and compare the resulting limits. Here the comparison is decisive: no single real number can be the limit from both sides.