Let $f: R \rightarrow R$ be a differential function with $f(2)=6$ and $f^{\prime}(2)=\frac{1}{48}$.

Question

Let $f: R \rightarrow R$ be a differential function with $f(2)=6$ and $f^{\prime}(2)=\frac{1}{48}$. Then $\lim _{x \rightarrow 2} \int_0^{f(x)} \frac{4 t^3}{x-2} d t=$

Professor PC Thomas Classes · Answer

L'Hopital's Rule + Leibniz rule: $\lim \frac{4(f(x))^3 \cdot f'(x)}{1} = 4(6)^3 (1/48) = 18$.