Let $$f : R \rightarrow R$$ be a function such that $$f(2) = 4$$ and $$f'(2) = 1$$. Then, the value of $$\lim_{x \to 2} \frac{x^2 f(2) - 4f(x)}{x - 2}$$ is equal to:

We have to evaluate the limit

$$\lim_{x\to 2}\frac{x^{2}\,f(2)-4\,f(x)}{x-2}.$$

The data given in the question are

$$f(2)=4\qquad\text{and}\qquad f'(2)=1.$$

First, substitute the value of $$f(2)$$ in the numerator:

$$x^{2}f(2)-4f(x)=x^{2}\cdot 4-4f(x)=4x^{2}-4f(x).$$

So the limit becomes

$$\lim_{x\to 2}\frac{4x^{2}-4f(x)}{x-2} =4\,\lim_{x\to 2}\frac{x^{2}-f(x)}{x-2}.$$

Now split the fraction in the bracket into two separate differences by adding and subtracting the same quantity 4 inside the numerator:

$$x^{2}-f(x)=\bigl(x^{2}-4\bigr)+\bigl(4-f(x)\bigr).$$

Hence,

$$\frac{x^{2}-f(x)}{x-2} =\frac{x^{2}-4}{x-2}+\frac{4-f(x)}{x-2}.$$

We shall evaluate the two limits that appear on the right separately.

First limit. Using the algebraic identity $$x^{2}-4=(x-2)(x+2),$$ we have

$$\frac{x^{2}-4}{x-2}=x+2,$$

which tends to

$$\lim_{x\to 2}(x+2)=4.$$

Second limit. Observe that

$$\frac{4-f(x)}{x-2}=-\frac{f(x)-4}{x-2}.$$

By the definition of the derivative,

$$\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}=f'(2)=1.$$

Replacing $$f(2)$$ by 4, we get

$$\lim_{x\to 2}\frac{f(x)-4}{x-2}=1.$$

Therefore,

$$\lim_{x\to 2}\frac{4-f(x)}{x-2}=-1.$$

Putting the two limits back together, we obtain

$$\lim_{x\to 2}\frac{x^{2}-f(x)}{x-2} =\;4+(-1)=3.$$

Finally, multiply by the factor 4 that we had taken out earlier:

$$\lim_{x\to 2}\frac{x^{2}f(2)-4f(x)}{x-2} =4\times 3=12.$$

Hence, the correct answer is Option D.