If $$f: R \rightarrow R$$ is a differentiable function and $$f(2) = 6$$, then $$\lim_{x \to 2} \int_6^{f(x)} \frac{2t \, dt}{(x - 2)}$$ is:

We have to evaluate the limit

$$L=\lim_{x \to 2}\,\frac{\displaystyle\int_{6}^{f(x)} 2t\,dt}{(x-2)}.$$

The numerator is a definite integral with a variable upper limit. To make the expression easier to handle, let us introduce a new function

$$F(x)=\int_{6}^{f(x)} 2t\,dt.$$

With this notation the required limit becomes

$$L=\lim_{x \to 2}\frac{F(x)}{x-2}.$$

Before applying the definition of the derivative, we must know the value of the function at the point $$x=2$$. Using the given information $$f(2)=6$$, we get

$$F(2)=\int_{6}^{f(2)} 2t\,dt=\int_{6}^{6} 2t\,dt=0.$$

Hence

$$L=\lim_{x \to 2}\frac{F(x)-F(2)}{x-2}.$$

But the limit on the right-hand side is exactly the first-principles (definition) formula of the derivative of $$F(x)$$ at $$x=2$$. Therefore

$$L=F'(2).$$

Now we must compute $$F'(x)$$. We recall the Leibniz Rule for differentiating an integral whose limit depends on $$x$$:

For a function $$G(x)=\int_{a}^{u(x)} g(t)\,dt$$ the derivative is given by $$G'(x)=g\!\big(u(x)\big)\,u'(x).$$

Here $$g(t)=2t$$ and $$u(x)=f(x)$$. Applying the rule gives

$$F'(x)=g\!\big(f(x)\big)\,f'(x)=\bigl(2f(x)\bigr)\,f'(x)=2\,f(x)\,f'(x).$$

We can now evaluate this derivative at $$x=2$$:

$$F'(2)=2\,f(2)\,f'(2).$$

The value $$f(2)$$ is provided in the question as $$f(2)=6,$$ so

$$F'(2)=2 \times 6 \times f'(2)=12\,f'(2).$$

Therefore the limit we were asked to find is

$$L=F'(2)=12\,f'(2).$$

Hence, the correct answer is Option D.