The third derivative of a function $f(x)$ vanishes for all $x$. If $f(0) = 1, f'(1) = 2$ and $f''(1) = -1$, then find $f''(x)$ at $x = 3$.
Step-by-Step Solution
Key Concept: For a limit of the form $\frac{\text{numerator}}{\text{denominator}}$ to exist as $x \to 0$, the leading coefficients must cancel, requiring specific parameter values.
The limit involves the ratio $\frac{a - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots - bx + cx^2 + x^3}{2x^3\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots\right) - 2x^3 + x^4}$. For the limit to exist, coefficients of $x^0$, $x^1$, and $x^2$ in the numerator must vanish: $a = b = 0$ and $c = \frac{a}{6}$. With these conditions, the limit equals $\frac{a}{120} \times \frac{3}{2} = \frac{3a}{240} = \frac{3}{40}$.
Correct Answer: 0