Why the Increment Cannot Be Set to Zero First
A difference quotient compares the change in a function’s output with a change in its input. The change in input must be nonzero: if the input does not change, the quotient asks us to divide by zero. The derivative is obtained not by setting that change equal to zero, but by taking a limit as the nonzero change approaches zero.
This distinction is easy to lose during algebra. A numerator may also become zero when the increment is zero, producing a form such as \(0/0\). That form is not a number and does not tell us the limit. The useful step is to simplify the quotient for nonzero increments, then evaluate the resulting limit.
The condition \(h\ne 0\) applies to the quotient before the limit is taken. The limit examines values for \(h\) close to zero but not equal to zero. Therefore, the quotient itself need not have a value at \(h=0\) for its limit to exist.
What \(0/0\) Does—and Does Not—Mean
Suppose direct substitution of \(h=0\) makes the numerator and denominator both zero. The resulting notation \(0/0\) is an indeterminate form: it is not the value of the quotient, and it does not determine the limit. Different quotients with this form can have different limits, or no limit at all. The original quotient is simply undefined at \(h=0\).
For a derivative calculation, the next question is whether algebra can simplify the quotient on its permitted domain \(h\ne 0\). For example, if the numerator has a factor \(h\), that factor may be canceled against the denominator because \(h\) is nonzero wherever the original quotient is defined. The simplified expression can then be used to find the limit.
Worked Example: The Derivative of \(x^2\) at \(3\)
Worked Example: The Derivative of \(x^2\) at \(3\)
Let \(f(x)=x^2\). Find \(f'(3)\) from the difference quotient.
Start with the definition. For \(h\ne 0\),
Expand the square and simplify the numerator:
Factor \(h\) in the numerator. Since the quotient is being considered only for \(h\ne 0\), cancel that factor:
Now take the limit as \(h\) approaches zero:
Check: For every nonzero \(h\), multiplying the simplified expression by the canceled denominator recovers the numerator: \(h(6+h)=6h+h^2\). The simplification is valid on the domain of the quotient, and the limit of \(6+h\) is \(6\). Thus \(f'(3)=6\).
Why Simplifying Before Taking the Limit Works
The difference quotient is the slope of a secant line through the points \((a,f(a))\) and \((a+h,f(a+h))\). A slope is change in output divided by change in input, so this secant slope is defined only when \(h\ne0\). As \(h\) approaches zero, the second point approaches the first. If the secant slopes approach a finite number, that number is the derivative at \(a\), the limiting tangent slope.
Algebraic simplification helps because the original formula may conceal how its values behave near zero. The justification is that equal expressions have equal values at every \(h\) in the original quotient’s domain. If a simplified expression agrees with the quotient for all sufficiently small nonzero \(h\), it has the same limit there. The value assigned at \(h=0\) is not needed to determine that limit.
Write that \(h\ne0\) while working with the difference quotient.
Expand, factor, or otherwise simplify without changing the quotient’s values for nonzero \(h\).
After simplification, let \(h\) approach zero. Do not substitute zero into the original quotient first.
Worked Example: A Zero Increment in the \(x\)-to-\(a\) Form
Worked Example: A Zero Increment in the \(x\)-to-\(a\) Form
Use the \(x\)-to-\(a\) difference quotient to find the derivative of \(f(x)=x^3\) at \(a=2\). In this form, \(x\) approaches \(a\), and the denominator requires \(x\ne a\).
Substituting \(x=2\) immediately would produce \(0/0\), but that is not a quotient value. Instead, factor the difference of cubes:
Now take the limit:
Check: Multiplying the canceled factor back gives \((x-2)(x^2+2x+4)=x^3-8\), so the factorization is correct. The limit of the simplified polynomial at \(2\) is \(12\), giving \(f'(2)=12\).
The two notations describe the same issue. In the \(h\)-form, the forbidden zero occurs at \(h=0\). In the \(x\)-to-\(a\) form, it occurs at \(x=a\). Replacing \(x\) with \(a+h\) makes the connection: \(x-a=h\). In either notation, the denominator is nonzero while the quotient is evaluated, and the derivative is a limit.
Cancellation Must Apply to Factors, Not Individual Terms
Another common error is canceling pieces of a sum or difference as though they were factors. Cancellation is valid for a common factor multiplying the entire numerator and denominator. It is not valid for a term that is only part of a sum.
Worked Example: Correcting an Invalid Cancellation
Consider \(f(x)=x^2\) at \(a=2\), using the \(x\)-to-\(a\) quotient:
A tempting but incorrect step is to “cancel the \(x\)” in \(x^2\) with the \(x\) in \(x-2\), producing \(\frac{x-4}{1-2}=4-x\). This is invalid: \(x^2-4\) is a difference of terms, not a product with a factor \(x\) common to the whole numerator and denominator.
Instead, factor the complete numerator as a difference of squares:
Therefore,
Check: Expanding the factored numerator gives \((x-2)(x+2)=x^2-4\). Also, the incorrect expression \(4-x\) would approach \(2\), not \(4\), so it does not give the correct derivative. Only the factorization that preserves the original quotient’s values for \(x\ne2\) is valid.
A Difference Quotient Is Not Yet the Derivative
Even when a difference quotient is correctly formed and evaluated, it is generally a number associated with a particular nonzero increment, not the derivative itself. The derivative is the limit of those quotient values as the increment tends to zero. A nearby secant slope can be close to the derivative without being equal to it.
Worked Example: A Nearby Quotient Versus the Derivative
For \(f(x)=x^2\) at \(a=2\), the quotient simplifies to \(4+h\) for \(h\ne0\). The derivative is the limit \(4\). Now use the particular increment \(h=0.1\):
This is the secant slope for an increment of \(0.1\), not the derivative. For \(h=-0.1\), the quotient is \(4+(-0.1)=3.9\). Taking the limit of \(4+h\) as \(h\) approaches zero gives \(4\).
Check: Substituting each nonzero increment into the simplified quotient gives the computed values: \(4+0.1=4.1\) and \(4-0.1=3.9\). Both are consistent with a limit of \(4\), but neither is itself that limit.
Common Errors and Their Corrections
| Error | Why it fails | Correction |
|---|---|---|
| Set \(h=0\) in the original quotient | The denominator becomes zero, so the quotient is undefined. | Simplify for \(h\ne0\), then take the limit. |
| Treat \(0/0\) as an answer | \(0/0\) is not a defined number and does not determine a limit. | Use valid algebra to find an expression for nearby nonzero increments. |
| Cancel terms across addition or subtraction | Cancellation requires a common factor of the entire numerator and denominator. | Factor first, then cancel only a common factor. |
| Call a quotient at one nonzero increment “the derivative” | That quotient is one secant slope; the derivative is the limiting value. | Keep the increment variable and evaluate its limit. |
A reliable habit is to write the restriction beside the quotient: \(h\ne0\), or \(x\ne a\). Keep it in view during any cancellation. Once the quotient has been simplified on that domain, take the limit. This prevents both the zero-denominator error and the mistaken belief that cancellation changes the original quotient’s domain.
Check Your Understanding
For each question, distinguish the quotient’s domain from the limit being evaluated.
- Why is \(\frac{f(a+h)-f(a)}{h}\) undefined at \(h=0\), even when its numerator is also zero?
- For \(f(x)=x^2\) at \(a=4\), simplify the difference quotient in \(h\)-form and find its limit as \(h\) approaches zero.
- In \(\frac{x^2-9}{x-3}\), why is canceling an \(x\) across the subtraction invalid? What factorization permits valid cancellation?
- For \(f(x)=x^2\) at \(a=1\), find the difference quotient for the specific increment \(h=0.2\). Is this value the derivative? Explain.
- State the difference between a difference quotient at a particular nonzero increment and the derivative at the point.
