Key Concepts and Formulas

• L'Hôpital's Rule: If $\mathop {\lim }\limits_{x \to c} \frac{g(x)}{h(x)}$ is of the indeterminate form $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{g(x)}{h(x)} = \mathop {\lim }\limits_{x \to c} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.
• Definition of the Derivative: The derivative of a function $f$ at a point $a$, denoted by $f'(a)$, is given by $f'(a) = \mathop {\lim }\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$.
• Limit Properties: The limit of a difference is the difference of the limits, and the limit of a product is the product of the limits, provided these limits exist.

Step-by-Step Solution

Step 1: Analyze the Indeterminate Form of the Limit We are asked to find the value of the limit: $L = \mathop {\lim }\limits_{x \to 2} \frac{x^2 f(2) - 4f(x)}{x - 2}$ Let's evaluate the numerator and the denominator as $x \to 2$. The denominator approaches $2 - 2 = 0$. The numerator approaches $(2)^2 f(2) - 4f(2) = 4f(2) - 4f(2) = 0$, since we are given $f(2) = 4$. Thus, the limit is of the indeterminate form $\frac{0}{0}$.

Step 2: Apply L'Hôpital's Rule Since the limit is in the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule. This involves taking the derivative of the numerator and the denominator with respect to $x$.

The numerator is $g(x) = x^2 f(2) - 4f(x)$. The derivative of the numerator is $g'(x) = \frac{d}{dx}(x^2 f(2) - 4f(x))$. Since $f(2)$ is a constant, its derivative is 0. The derivative of $x^2 f(2)$ with respect to $x$ is $2x f(2)$. The derivative of $-4f(x)$ with respect to $x$ is $-4f'(x)$. So, $g'(x) = 2x f(2) - 4f'(x)$.

The denominator is $h(x) = x - 2$. The derivative of the denominator is $h'(x) = \frac{d}{dx}(x - 2) = 1$.

Applying L'Hôpital's Rule, the limit becomes: $L = \mathop {\lim }\limits_{x \to 2} \frac{g'(x)}{h'(x)} = \mathop {\lim }\limits_{x \to 2} \frac{2x f(2) - 4f'(x)}{1}$

Step 3: Evaluate the Limit using the given values Now we substitute the values $x = 2$, $f(2) = 4$, and $f'(2) = 1$ into the expression obtained in Step 2. $L = 2(2) f(2) - 4f'(2)$ $L = 4 f(2) - 4f'(2)$ Substitute the given values $f(2) = 4$ and $f'(2) = 1$: $L = 4(4) - 4(1)$ $L = 16 - 4$ $L = 12$

Alternative Approach (without direct L'Hôpital's Rule on the entire expression):

Step 1 (Alternative): Rearrange the Limit Expression We can rewrite the limit expression by adding and subtracting terms in the numerator to create a form related to the definition of the derivative. $L = \mathop {\lim }\limits_{x \to 2} \frac{x^2 f(2) - 4f(x)}{x - 2}$ We know $f(2) = 4$. Substitute this into the numerator: $L = \mathop {\lim }\limits_{x \to 2} \frac{x^2 (4) - 4f(x)}{x - 2}$ $L = \mathop {\lim }\limits_{x \to 2} \frac{4x^2 - 4f(x)}{x - 2}$ We can factor out 4 from the numerator: $L = 4 \mathop {\lim }\limits_{x \to 2} \frac{x^2 - f(x)}{x - 2}$ Now, let's manipulate the numerator $x^2 - f(x)$ to involve $f(2)$. We can add and subtract $4x$ to create terms related to $f(x) - f(2)$: $L = 4 \mathop {\lim }\limits_{x \to 2} \frac{x^2 - 4x + 4x - f(x)}{x - 2}$ $L = 4 \mathop {\lim }\limits_{x \to 2} \frac{(x^2 - 4x) - (f(x) - 4x)}{x - 2}$ This doesn't seem to directly lead to the derivative of $f(x)$. Let's try a different manipulation.

Let's go back to: $L = \mathop {\lim }\limits_{x \to 2} \frac{x^2 f(2) - 4f(x)}{x - 2}$ Substitute $f(2) = 4$: $L = \mathop {\lim }\limits_{x \to 2} \frac{4x^2 - 4f(x)}{x - 2}$ We want to create terms of the form $f(x) - f(2)$. Let's add and subtract $4 \cdot 2^2 = 16$ in the numerator and also $4 \cdot 2 = 8$ in a way that helps. $L = \mathop {\lim }\limits_{x \to 2} \frac{4x^2 - 16 - 4f(x) + 16}{x - 2}$ $L = \mathop {\lim }\limits_{x \to 2} \frac{4(x^2 - 4) - 4(f(x) - 4)}{x - 2}$ $L = \mathop {\lim }\limits_{x \to 2} \left( \frac{4(x^2 - 4)}{x - 2} - \frac{4(f(x) - 4)}{x - 2} \right)$ We can split this into two limits: $L = 4 \mathop {\lim }\limits_{x \to 2} \frac{x^2 - 4}{x - 2} - 4 \mathop {\lim }\limits_{x \to 2} \frac{f(x) - 4}{x - 2}$

Step 2 (Alternative): Evaluate the first limit The first limit is: $\mathop {\lim }\limits_{x \to 2} \frac{x^2 - 4}{x - 2}$ This is a difference of squares in the numerator: $x^2 - 4 = (x - 2)(x + 2)$. $\mathop {\lim }\limits_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2}$ For $x \neq 2$, we can cancel out $(x - 2)$: $\mathop {\lim }\limits_{x \to 2} (x + 2) = 2 + 2 = 4$

Step 3 (Alternative): Evaluate the second limit The second limit is: $\mathop {\lim }\limits_{x \to 2} \frac{f(x) - 4}{x - 2}$ We are given that $f(2) = 4$. So, we can replace 4 with $f(2)$: $\mathop {\lim }\limits_{x \to 2} \frac{f(x) - f(2)}{x - 2}$ This is precisely the definition of the derivative of $f$ at $x = 2$, which is $f'(2)$. We are given $f'(2) = 1$.

Step 4 (Alternative): Combine the results Now, substitute the values of the two limits back into the expression for $L$: $L = 4 \times (\text{value of first limit}) - 4 \times (\text{value of second limit})$ $L = 4 \times (4) - 4 \times (1)$ $L = 16 - 4$ $L = 12$

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$) before applying the rule.
• Errors in Differentiation: Be careful when differentiating expressions involving $f(x)$ and $f'(x)$. Remember that $f(2)$ is a constant.
• Algebraic Manipulation Errors: When using the alternative method, ensure that the algebraic steps to create terms related to the definition of the derivative are correct.

Summary

The problem asks for the evaluation of a limit that is initially in an indeterminate form $\frac{0}{0}$. We can solve this using L'Hôpital's Rule by differentiating the numerator and the denominator separately and then evaluating the resulting expression at $x=2$, using the given values $f(2)=4$ and $f'(2)=1$. Alternatively, we can manipulate the limit expression algebraically to isolate terms that represent the definition of the derivative of $f$ at $x=2$ and a standard limit of a polynomial. Both methods lead to the same result.

The final answer is $\boxed{12}$.