Question

Let $F:R\to R$ be a differentiable function such that $F\left(x\right)={\int }_4^{f\left(x\right)}\frac{16t^3}{x-3}dt,f\left(3\right)=4\&f^{\prime }\left(3\right)=\frac{1}{16}$ then the value of $\underset{x\to 3}{lim}F\left(x\right)$ is

Answer 1

$F\left(x\right)={\int }_4^{f\left(x\right)}\frac{16t^3}{x-3}dt$
$\underset{x\to 3}{lim}F\left(x\right)=\underset{x\to 3}{lim}\frac{\underset{4}{\overset{f\left(x\right)}{\int }}16t^3}{x-3}dt=\underset{x\to 3}{lim}\frac{16\left(f\right(x\left){)}^3{f}^{\prime }\right(x)}{1}(\frac{0}{0},\text{L-Hospital})=\frac{16\times 4^3}{1}\times \frac{1}{16}=64$