We are given that a function $f(x)$ satisfies two conditions:

1. $f(a + b) = f(a) + f(b) + ab$ for all real numbers $a$ and $b$.
2. $\lim_{h \to 0} \frac{f(h)}{h} = 0$.

We are asked to find $f'(x)$ using the definition of the derivative. Let's proceed step-by-step.

Step 1: Use the derivative definition

The derivative of $f(x)$ at a point $x$ is defined as:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.$

To apply this, we first need to evaluate $f(x + h) - f(x)$. Using the given functional equation:

$f(x + h) = f(x) + f(h) + xh.$

Thus, we can express $f(x + h) - f(x)$ as:

$f(x + h) - f(x) = (f(x) + f(h) + xh) - f(x) = f(h) + xh.$

Step 2: Substitute into the derivative formula

Now we substitute this expression for $f(x + h) - f(x)$ into the derivative formula:

$f'(x) = \lim_{h \to 0} \frac{f(h) + xh}{h}.$

We can separate the terms in the numerator:

$f'(x) = \lim_{h \to 0} \left( \frac{f(h)}{h} + x \right).$

Step 3: Use the given limit

We are given that $\lim_{h \to 0} \frac{f(h)}{h} = 0$. So, applying this in the above expression:

$f'(x) = 0 + x = x.$

Final Answer:

Thus, the derivative of $f(x)$ is:

$f'(x) = x.$

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Would you like further details on this solution, or do you have any questions?

Here are five related questions for further exploration:

1. How does the functional equation $f(a + b) = f(a) + f(b) + ab$ lead to $f(x) = \frac{x^2}{2}$ as a possible solution?
2. How can we verify that the function $f(x) = \frac{x^2}{2}$ satisfies the given functional equation?
3. What is the significance of the limit condition $\lim_{h \to 0} \frac{f(h)}{h} = 0$?
4. How would the derivative change if the functional equation was $f(a + b) = f(a) + f(b)$ instead?
5. Can we generalize the solution to find derivatives for other similar functions?

Tip: When using the derivative definition, always look for ways to manipulate the functional equation into a form where you can apply the given limits or other properties.