Let $$f(x)$$ be a polynomial function such that $$f(x) + f'(x) + f''(x) = x^5 + 64$$. Then, the value of $$\lim_{x \to 1} \frac{f(x)}{x - 1}$$ is equal to

We need to find $$\lim_{x \to 1} \frac{f(x)}{x - 1}$$ where $$f(x) + f'(x) + f''(x) = x^5 + 64$$. Since the right-hand side is a degree 5 polynomial, it follows that $$f(x)$$ must also be of degree 5; accordingly, let $$f(x) = x^5 + a x^4 + b x^3 + c x^2 + d x + e$$ so that $$f'(x) = 5x^4 + 4a x^3 + 3b x^2 + 2c x + d$$ and $$f''(x) = 20x^3 + 12a x^2 + 6b x + 2c$$.

By equating coefficients of like powers of $$x$$ one obtains:
x^5: 1 = 1 ✓
x^4: a + 5 = 0 ⇒ a = -5
x^3: b + 4a + 20 = 0 ⇒ b + 4(-5) + 20 = 0 ⇒ b = 0
x^2: c + 3b + 12a = 0 ⇒ c + 0 - 60 = 0 ⇒ c = 60
x^1: d + 2c + 6b = 0 ⇒ d + 120 + 0 = 0 ⇒ d = -120
x^0: e + d + 2c = 64 ⇒ e - 120 + 120 = 64 ⇒ e = 64

Hence $$f(x) = x^5 - 5x^4 + 60x^2 - 120x + 64$$, and substituting $$x = 1$$ gives $$f(1) = 1 - 5 + 60 - 120 + 64 = 0$$, so the limit is of the form $$\frac{0}{0}$$. Thus one can apply L'Hôpital's rule or observe that $$\lim_{x \to 1} \frac{f(x)}{x - 1} = f'(1)$$.

Since $$f'(x) = 5x^4 - 20x^3 + 120x - 120$$, one finds $$f'(1) = 5 - 20 + 120 - 120 = -15$$.

Option A: $$-15$$.