Let and be twice-differentiable functions such that

- If , compute .
- If , compute .
- Compute .

- Using the rule for the derivative of a quotient we have,
Substituting the given expressions for , and we have,

- Using the product rule (more precisely, the product rule for more than two terms that we proved in this exercise) we have,
Then, we evaluate at 0,

since we are given .

- Since we are given that and are twice differentiable functions, we know that and are differentiable; hence, and are continuous since differentiability implies continuity (from Section 4.4 of Apostol). From the definition of continuity of a function at a point we know this means,
Furthermore, since and . So we can then compute,

since we are given .