How to find $\lim_{n\to\infty}\int_{1/n}^{2/n}\frac{f(x)}{x}\,\mathrm{d}x$? 
Suppose $f:\mathbf{R}\to\mathbf{R}$ is continuous and $f(0)=3$. Find $$\lim_{n\to\infty}\int_{1/n}^{2/n}\frac{f(x)}{x}\mathrm{d}x.$$

I don't know how to do this problem, any hints?
 A: $f$ is continuous at $0$, hence there exists $\delta>0$ such that $|f(x)-3|<\delta$ if $|x|<\varepsilon$. It follows that the integral $L$ is such that
$$
(3-\delta)\log 2=\int_{1/n}^{2/n}\frac{3-\delta}{x}\mathrm{d}x\le L\le \int_{1/n}^{2/n}\frac{3+\delta}{x}\mathrm{d}x=(3+\delta)\log 2
$$
if $n$ is sufficiently large. Hence the limit is $3\log 2$.
(Thanks to Michael Hardy and Winther for spotting a mistake.)
A: (Too slow to finish this. Well, I agree with @Paolo.)
Given $\epsilon>0$, by continuity of $f$ at $x=0$, 
$$3-\epsilon<f(x)<3+\epsilon$$ for all $x\in[\frac{1}{n},\frac{2}{n}]$ for large enough $n$ and thus
$$
(3-\epsilon)\int_{1/n}^{2/n}\frac{1}{x}\,dx\leq \int_{1/n}^{2/n}\frac{f(x)}{x}\,dx\leq (3+\epsilon)\int_{1/n}^{2/n}\frac{1}{x}\,dx
$$
which implies that for large enough $n$,
$$
(3-\epsilon)\ln 2\leq \int_{1/n}^{2/n}\frac{f(x)}{x}\,dx\leq (3+\epsilon)\ln 2.
$$
It follows that 
$$
(3-\epsilon)\ln 2\leq \liminf_nL_n\leq\limsup_nL_n\leq (3+\epsilon)\ln 2
$$
where $L_n$ denotes the integral in question. 
Since $\epsilon$ is arbitrary, one must have
$$
3\ln 2\leq \liminf_nL_n\leq\limsup_nL_n\leq 3\ln 2
$$
which implies $\lim_nL_n=3\ln 2$.
A: Just another idea (not simpler or more elementary though). 
Set $u=nx$. Then integral becomes 
$$ \int_1^2 \frac{f(u/n)}{u} du.$$ 
By continuity of $f$ at $0$, the integrand converges uniformly to $f(0)/u$. By continuity of integral with respect to uniform convergence,  the limit of the integral is  $f(0) (\ln 2-\ln 1)=3 \ln 2$.  
A: According to the mean value theorem for integrals (see the second formulation at this wiki page) we have
\begin{equation*}
\int_{1/n}^{2/n}\dfrac{f(x)}{x}\, dx = f(\xi_{n}) \int_{1/n}^{2/n}\dfrac{1}{x}\, dx = f(\xi_{n})\ln 2 \to f(0)\ln 2 = 3\ln 2,\quad n \to \infty
\end{equation*}
where $1/n < \xi_{n} <2/n $.
