$\lim_{n \to \infty}\frac{1}{(1/n)}\int^{b+\frac{1}{n}}_{b} F(x)dx=F(b)$ Suppose we are given a continuous function $F$ which is also increasing and bounded. I was reading a proof which involved asserting that 

$\lim_{n \to \infty}\frac{1}{(1/n)}\int^{b+\frac{1}{n}}_{b} F(x) dx=F(b)$

I am actually having some trouble seeing this. I think this should just be an application of L'Hopital's since the limit has the indeterminate form 
$\frac{0}{0}$ since the denominator $\frac{1}{n} \to 0$ and the numerator $\int^{b+\frac{1}{n}}_{b} F(x) dx \to 0$ as $n \to \infty$.
So in my botched attempt at L'Hopital's rule, I use the Fundamental Theorem of Calculus to get $F(b+\frac{1}{n})$ in the numerator and $\frac{-1}{n^{2}}$ in the denominator. I must have did something wrong or must not be looking at this the right way. 
 A: This is a consequence of the Fundamental Theorem of Calculus:\begin{align}\left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0}\int_b^{b+t}F(x)\,\mathrm dx=F(b)&\iff\lim_{h\to0}\frac{\int_b^{b+h}F(x)\,\mathrm dx}h\\&\implies\lim_{n\to\infty}\frac{\int_b^{b+\frac1n}F(x)\,\mathrm dx}{\frac1n}.\end{align}
A: Write $n\int_b^{b+1/n} F(x) dx = n\int_b^{b+1/n}(F(x) - F(b)) dx+F(b).$ Since $F$ is continuous at $b,$ $F(x)-F(b)$ will be arbitrarily small, hence, the integral too. QED
A: What you've neglected in your attempt to use L'Hopital is to use the Chain Rule  in addition to the Fundamental Theorem of Calculus when taking the derivative of the integral:  When you apply L'Hopital to $\lim_{n\to\infty}{f(n)\over g(n)}$ with $f(n)=\int_b^{b+1/n}F(x)dx$ and $g(n)=1/n$, you get
$$f'(n)=F\left(b+{1\over n}\right)\left(1\over n\right)'=F\left(b+{1\over n}\right){1\over n^2}$$
and the $1/n^2$ cancels the $g'(n)=1/n^2$, leaving just $F(b+1/n)$, which tends to $F(b)$ as $n\to\infty$ (since $F$ is assumed to be continuous).
