Let $f:\mathbb{R}\to \mathbb{R}$ be a differentiable function having $f(2)=6,f^{\prime }(2)=\frac{1}{48}$. Then ${\lim }_{x\to 2}\frac{{\int }_6^{f(x)}4t^{3}dt}{x-2}$ equals:

The limit is of the form $0/0$ since $f(2)=6$ and ${\int }_6^{6}4t^{3}dt=0$.Applying L'Hospital's Rule:
$$\underset{x\to 2}{\lim }\frac{\frac{d}{dx}{\int }_6^{f(x)}4t^{3}dt}{\frac{d}{dx}(x-2)}$$
Using Leibniz's Rule for the numerator:
$$\underset{x\to 2}{\lim }\frac{4[f(x){]}^3\cdot f^{\prime }(x)}{1}$$
Substituting $x=2$:
$$4[f(2){]}^3\cdot f^{\prime }(2)=4(6{)}^3\cdot \frac{1}{48}$$
$$=4\cdot 216\cdot \frac{1}{48}=\frac{864}{48}=18$$