1. The problem is to understand the limit expression for the derivative: $$f'(x) = \lim_{h \to 0} \frac{12xh + 6h^2}{h}$$ 2. We factor out the common factor $h$ from the numerator: $$12xh + 6h^2 = h(12x + 6h)$$ 3. Substituting back into the limit expression: $$f'(x) = \lim_{h \to 0} \frac{h(12x + 6h)}{h}$$ 4. Since $h \neq 0$ in the limit process (we approach zero but never equal zero), we can cancel $h$ in numerator and denominator: $$f'(x) = \lim_{h \to 0} (12x + 6h)$$ 5. Now, as $h$ approaches 0, the term $6h$ approaches 0, so: $$f'(x) = 12x + 0 = 12x$$ 6. The $6$ did not disappear; it is still multiplied by $h$ inside the parentheses. When $h$ goes to zero, $6h$ goes to zero, so it vanishes in the limit. In summary, factoring out $h$ keeps the $6$ inside the parentheses, and it only disappears after taking the limit as $h \to 0$ because $6h$ becomes zero.