How to calculate the derivative of $\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ at $x=0$? Let $F(x):=\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ ,where $[\frac{1}{t}]$ is the largest integer no more than $\frac{1}{t}$.Prove $F'(0)=\frac{1}{2}$.
I have tried in this way:
\begin{equation}
\begin{aligned}
\lim_{n\to\infty}nF\left(\frac{1}{n}\right)&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-\left[\frac{1}{t}\right]\right)dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-k\right) dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\ln(1+\frac{1}{k})-\frac{1}{k+1}\right)\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\frac{1}{k}-\frac{1}{k+1}\right)\\&=1.
\end{aligned}
\end{equation}
Please give me some ideas,thank you!

i) Consider $$ A_n :=\int^{\frac{1}{n-1}}_{ \frac{1}{n} } \
\frac{1}{x}\ dx - (n-1) \bigg[ \frac{1}{n-1} - \frac{1}{n}\bigg] $$
Hence $F(\frac{1}{n-1})=A_n +A_{n+1}+\cdots $
So $$ A_n\bigg[ \frac{1}{n-1} - \frac{1}{n}\bigg]^{-1} \rightarrow
\frac{1}{2}$$
so that $\frac{F(\frac{1}{n-1}) }{ \frac{1}{n-1} } \rightarrow
\frac{1}{2} $
ii) $ \frac{F(\frac{1}{n} )}{\frac{1}{n-1}}\leq \frac{F(x)}{x} \leq
\frac{F(\frac{1}{n-1} )}{\frac{1}{n}} $ where $\frac{1}{n}\leq
x<\frac{1}{n-1} $ 
