How can the limit definition calculate a derivative?

It asks how the function’s output changes, on average, between x=a and a nearby input a+h. The difference quotient is [f(a+h)−f(a)]/h. The derivative at a is the limit of that quotient as h approaches 0, provided the limit exists. [S1]

The order matters: first make and simplify the quotient for h≠0; then take the limit. You are not asked to plug h=0 into the unsimplified fraction, which would make its denominator zero. The limit describes what the quotient approaches for nearby nonzero h.

Start by substituting 3+h and 3 into the function:

1. f(3+h)=(3+h)²=9+6h+h², and f(3)=9.
2. Put these into the difference quotient:

[(9+6h+h²)−9]/h.

1. Combine like terms and factor out h:

(6h+h²)/h = h(6+h)/h.

1. For h≠0, cancel h. The quotient is now 6+h.
2. Take the limit as h approaches 0: 6+h approaches 6.

So the derivative at x=3 is 6. The key limiting step is the last one: the derivative is the limit of the simplified quotient. The cancellation is valid for nonzero h near 0; it does not mean the original quotient was defined at h=0.

Because the quotient has h in its denominator. Direct substitution gives division by zero, so it does not give a value for the quotient at h=0. Instead, simplify for h≠0 and ask what the simplified expression approaches as h gets closer to zero.

In this example, cancellation turns the quotient into 6+h wherever h≠0. As h approaches 0, that expression approaches 6. This is precisely the distinction in the definition: the derivative is the limit, not a value obtained by putting zero into the original quotient. [S1] The method gives a derivative at the chosen point only if the relevant limit exists; this example’s algebra shows that it does.

Use this template for the derivative at x=a:

1. Write the definition: f′(a)=lim as h→0 of [f(a+h)−f(a)]/h. [S1]
2. Calculate f(a+h) and f(a) separately.
3. Substitute both results into the quotient, using parentheses to preserve subtraction.
4. Simplify and factor; cancel h only when it is a common factor and h≠0.
5. Take the limit of what remains as h→0, and state the result only if that limit exists.

For the worked example, the calculation is: f(x)=x², a=3; [f(3+h)−f(3)]/h = [(3+h)²−9]/h = (6h+h²)/h = 6+h for h≠0; therefore lim as h→0 (6+h)=6, so f′(3)=6.

For a new function, keep the same structure but redo the substitution and simplification. If the quotient does not simplify neatly, check the algebra before concluding that a derivative does or does not exist.