Key Concepts and Formulas

• Polynomial Properties: If $P(x)$ is a polynomial, then $P(x) + P'(x) + P''(x)$ will also be a polynomial. The degree of $P(x) + P'(x) + P''(x)$ is the same as the degree of $P(x)$.
• Equating Coefficients: For two polynomials to be equal for all values of $x$, the coefficients of corresponding powers of $x$ must be equal.
• L'Hôpital's Rule: If a limit of a function results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, L'Hôpital's Rule can be applied by taking the derivative of the numerator and the denominator separately and then evaluating the limit.
• Limit Definition of the Derivative: The limit $\mathop {\lim }\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$ represents the derivative of the function $f(x)$ at $x=a$, denoted as $f'(a)$.

Step-by-Step Solution

Step 1: Determine the degree of the polynomial $f(x)$. We are given the equation $f(x) + f'(x) + f''(x) = x^5 + 64$. Since $f(x)$ is a polynomial, $f'(x)$ and $f''(x)$ are also polynomials. The degree of $f'(x)$ is one less than the degree of $f(x)$, and the degree of $f''(x)$ is two less than the degree of $f(x)$. The sum $f(x) + f'(x) + f''(x)$ will have the same degree as $f(x)$ because the highest degree terms will come from $f(x)$ itself. The right-hand side of the equation is $x^5 + 64$, which is a polynomial of degree 5. Therefore, $f(x)$ must be a polynomial of degree 5.

Step 2: Assume a general form for $f(x)$ and its derivatives. Let $f(x) = Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F$. Then, its first derivative is $f'(x) = 5Ax^4 + 4Bx^3 + 3Cx^2 + 2Dx + E$. And its second derivative is $f''(x) = 20Ax^3 + 12Bx^2 + 6Cx + 2D$.

Step 3: Substitute the expressions for $f(x)$, $f'(x)$, and $f''(x)$ into the given equation. $(Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F) + (5Ax^4 + 4Bx^3 + 3Cx^2 + 2Dx + E) + (20Ax^3 + 12Bx^2 + 6Cx + 2D) = x^5 + 64$

Step 4: Group terms by powers of $x$ on the left-hand side. $Ax^5 + (B+5A)x^4 + (C+4B+20A)x^3 + (D+3C+12B)x^2 + (E+2D+6C)x + (F+E+2D) = x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 64$

Step 5: Equate the coefficients of corresponding powers of $x$ on both sides of the equation. By comparing the coefficients of $x^5$: $A = 1$

By comparing the coefficients of $x^4$: $B + 5A = 0$ $B + 5(1) = 0 \Rightarrow B = -5$

By comparing the coefficients of $x^3$: $C + 4B + 20A = 0$ $C + 4(-5) + 20(1) = 0$ $C - 20 + 20 = 0 \Rightarrow C = 0$

By comparing the coefficients of $x^2$: $D + 3C + 12B = 0$ $D + 3(0) + 12(-5) = 0$ $D - 60 = 0 \Rightarrow D = 60$

By comparing the coefficients of $x$: $E + 2D + 6C = 0$ $E + 2(60) + 6(0) = 0$ $E + 120 = 0 \Rightarrow E = -120$

By comparing the constant terms: $F + E + 2D = 64$ $F + (-120) + 2(60) = 64$ $F - 120 + 120 = 64 \Rightarrow F = 64$

Step 6: Write down the polynomial $f(x)$ using the determined coefficients. Using the coefficients $A=1, B=-5, C=0, D=60, E=-120, F=64$, we get: $f(x) = x^5 - 5x^4 + 0x^3 + 60x^2 - 120x + 64$ $f(x) = x^5 - 5x^4 + 60x^2 - 120x + 64$

Step 7: Evaluate the limit $\mathop {\lim }\limits_{x \to 1} \frac{f(x)}{x - 1}$. We need to find $\mathop {\lim }\limits_{x \to 1} \frac{x^5 - 5x^4 + 60x^2 - 120x + 64}{x - 1}$. Let's check the value of the numerator at $x=1$: $f(1) = (1)^5 - 5(1)^4 + 60(1)^2 - 120(1) + 64 = 1 - 5 + 60 - 120 + 64 = 125 - 125 = 0$. Since the numerator is 0 and the denominator is $1-1=0$ at $x=1$, this is an indeterminate form of type $\frac{0}{0}$.

Step 8: Apply L'Hôpital's Rule to evaluate the limit. According to L'Hôpital's Rule, we can take the derivative of the numerator and the denominator separately. Derivative of the numerator: $\frac{d}{dx}(x^5 - 5x^4 + 60x^2 - 120x + 64) = 5x^4 - 20x^3 + 120x - 120$. Derivative of the denominator: $\frac{d}{dx}(x - 1) = 1$.

Now, we evaluate the limit of the ratio of these derivatives: $\mathop {\lim }\limits_{x \to 1} \frac{5x^4 - 20x^3 + 120x - 120}{1}$ Substitute $x=1$ into this expression: $5(1)^4 - 20(1)^3 + 120(1) - 120 = 5 - 20 + 120 - 120 = 5 - 20 = -15$.

Alternatively, we can recognize that $\mathop {\lim }\limits_{x \to 1} \frac{f(x)}{x - 1}$ is the definition of $f'(1)$ if $f(1)=0$. From Step 6, we found $f(x) = x^5 - 5x^4 + 60x^2 - 120x + 64$. We calculated $f(1)=0$. Now we find $f'(x)$: $f'(x) = 5x^4 - 20x^3 + 120x - 120$. Evaluating $f'(1)$: $f'(1) = 5(1)^4 - 20(1)^3 + 120(1) - 120 = 5 - 20 + 120 - 120 = -15$.

Common Mistakes & Tips

• Incorrect Degree Assumption: Assuming a wrong degree for $f(x)$ will lead to incorrect coefficients. Always verify that the degree of $f(x)$ matches the degree of the resulting polynomial sum.
• Algebraic Errors in Coefficient Matching: Carefully equate coefficients to avoid calculation mistakes. A single error can propagate through the entire solution.
• L'Hôpital's Rule Application: Ensure the limit is in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying L'Hôpital's Rule. Also, remember to differentiate the numerator and denominator separately, not the entire fraction.
• Recognizing the Derivative Definition: The limit $\mathop {\lim }\limits_{x \to 1} \frac{f(x)}{x - 1}$ is equivalent to $f'(1)$ if $f(1)=0$. This can sometimes be a quicker way to evaluate the limit after finding $f(x)$.

Summary

The problem requires finding the value of a limit involving a polynomial function $f(x)$ that satisfies a differential equation. We first determined the degree of the polynomial $f(x)$ by comparing it with the given equation. Then, we assumed a general form for $f(x)$, calculated its first and second derivatives, and substituted them into the given equation. By equating the coefficients of corresponding powers of $x$, we found the specific coefficients of $f(x)$. Finally, we evaluated the required limit. Since the limit resulted in an indeterminate form, L'Hôpital's Rule was applied, or alternatively, the limit was recognized as the derivative of $f(x)$ at $x=1$.

The value of the limit is -15.

The final answer is \boxed{-15}.