Key Concepts and Formulas

• Definition of the Derivative: The derivative of a function $f(x)$ at a point $a$, denoted by $f'(a)$, is defined as: $f'(a) = \mathop {\lim }\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$
• Limit Properties: The limit of a sum is the sum of the limits, and the limit of a constant times a function is the constant times the limit of the function, provided these limits exist.
  • $\mathop {\lim }\limits_{x \to a} [g(x) + h(x)] = \mathop {\lim }\limits_{x \to a} g(x) + \mathop {\lim }\limits_{x \to a} h(x)$
  • $\mathop {\lim }\limits_{x \to a} [c \cdot g(x)] = c \cdot \mathop {\lim }\limits_{x \to a} g(x)$

Step-by-Step Solution

We are asked to evaluate the limit: $L = \mathop {\lim }\limits_{x \to 2} \frac{xf(2) - 2f(x)}{x - 2}$

Step 1: Manipulate the numerator to introduce terms involving $f(2)$ and $f(x) - f(2)$. To apply the definition of the derivative, we need to create terms of the form $f(x) - f(2)$ in the numerator. We can achieve this by adding and subtracting $2f(2)$ in the numerator. $L = \mathop {\lim }\limits_{x \to 2} \frac{xf(2) - 2f(2) + 2f(2) - 2f(x)}{x - 2}$

Step 2: Group terms to facilitate factorization and application of the derivative definition. Group the first two terms and the last two terms in the numerator. $L = \mathop {\lim }\limits_{x \to 2} \frac{[xf(2) - 2f(2)] - [2f(x) - 2f(2)]}{x - 2}$

Step 3: Factor out common terms from each group in the numerator. Factor $f(2)$ from the first group and $2$ from the second group. $L = \mathop {\lim }\limits_{x \to 2} \frac{f(2)(x - 2) - 2[f(x) - f(2)]}{x - 2}$

Step 4: Split the fraction into two separate fractions. This allows us to evaluate the limit of each part independently. $L = \mathop {\lim }\limits_{x \to 2} \left( \frac{f(2)(x - 2)}{x - 2} - \frac{2[f(x) - f(2)]}{x - 2} \right)$

Step 5: Simplify the first term and apply limit properties to the second term. The first term simplifies to $f(2)$, as $x \to 2$ implies $x \neq 2$. For the second term, we can pull out the constant factor of $2$. $L = \mathop {\lim }\limits_{x \to 2} f(2) - 2 \mathop {\lim }\limits_{x \to 2} \frac{f(x) - f(2)}{x - 2}$

Step 6: Evaluate the limits. The limit of a constant $f(2)$ as $x \to 2$ is simply $f(2)$. The second limit is the definition of the derivative of $f$ at $x=2$, which is $f'(2)$. $L = f(2) - 2 f'(2)$

Step 7: Substitute the given values of $f(2)$ and $f'(x)$. We are given that $f(2) = 4$ and $f'(x) = 4$ for all $x$. Therefore, $f'(2) = 4$. $L = 4 - 2(4)$

Step 8: Calculate the final result. $L = 4 - 8$ $L = -4$

Common Mistakes & Tips

• Incorrectly applying the derivative definition: Ensure the numerator has the form $f(x) - f(a)$ and the denominator has $x-a$. Manipulating the numerator by adding and subtracting terms is a common technique.
• Algebraic errors: Be careful with signs when rearranging and factoring terms in the numerator.
• Assuming $f'(x)$ is constant: The problem states $f'(x) = 4$. This implies that the derivative is constant for all $x$, so $f'(2) = 4$. If the derivative was given as a function of $x$, say $f'(x) = g(x)$, then the limit would be $f(2) - 2g(2)$.

Summary

The problem requires evaluating a limit that resembles the definition of a derivative. By strategically adding and subtracting $2f(2)$ in the numerator, we were able to rewrite the expression as a difference of two limits. The first limit simplified to $f(2)$, and the second limit, after factoring out a constant, became $f'(2)$. Substituting the given values of $f(2)$ and $f'(x)$ allowed us to compute the final numerical answer.

The final answer is $\boxed{-4}$.