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Step-by-Step Solution
Step 1: Identify the Given Information
We have a function $f: \mathbb{R} \to \mathbb{R}$ with:
$f(2) = 4$
$f'(2) = 1$
We need to evaluate the limit
$ \displaystyle \lim_{x \to 2} \frac{x^2 \, f(2) - 4\, f(x)}{x - 2} $.
Step 2: Check the Indeterminate Form
First, substitute $x = 2$ into the numerator:
\[
x^2 f(2) - 4\,f(x)\bigg|_{x=2} \;=\; 2^2 \cdot f(2) \;-\; 4 f(2)
\;=\; 4 \cdot 4 \;-\; 4 \cdot 4
\;=\; 16 \;-\; 16
\;=\; 0.
\]
The denominator $x - 2$ also becomes $0$ when $x = 2$. Hence, the expression is of the form $\frac{0}{0}$, an indeterminate form.
Step 3: Apply L'Hôpital's Rule
Because we have an indeterminate form $\tfrac{0}{0}$, we can apply L'Hôpital's Rule, which tells us to take the derivative of the numerator and the derivative of the denominator with respect to $x$, and then find the limit.
3.1: Differentiate the Numerator
Let $N(x) = x^2 \, f(2) - 4\,f(x).$
Since $f(2)$ is a constant (equals 4), its derivative is zero with respect to $x$, while $f(x)$ differentiates to $f'(x)$. We get:
\[
N'(x)
= \frac{d}{dx}\bigl(x^2 \cdot f(2)\bigr)
- 4\,\frac{d}{dx}\bigl(f(x)\bigr).
\]
\[
= 2x \, f(2) \;-\; 4\,f'(x).
\]
3.2: Differentiate the Denominator
The denominator is $D(x) = x - 2$. Its derivative is:
\[
D'(x) = \frac{d}{dx}(x - 2) = 1.
\]
Step 4: Evaluate the Limit Using the Derivatives
Now, by L’Hôpital’s Rule:
\[
\lim_{x \to 2} \frac{x^2 \, f(2) - 4\, f(x)}{x - 2}
\;=\;
\lim_{x \to 2} \frac{N'(x)}{D'(x)}
\;=\;
\lim_{x \to 2} \bigl(2x \, f(2) - 4 \, f'(x)\bigr).
\]
Substitute $x = 2$, $f(2) = 4$, and $f'(2) = 1$:
\[
2 \cdot 2 \cdot 4
\;-\; 4 \cdot f'(2)
\;=\; 16 \;-\; 4 \cdot 1
\;=\; 16 \;-\; 4
\;=\; 12.
\]
Step 5: State the Final Answer
The value of the given limit is $\boxed{12}$.
