Undergraduate math competition problem: find $\lim \limits_{n \to \infty} \int \limits^{2006}_{1385}f(nx)\, \mathrm dx$ Suppose $f\colon [0, +\infty) \to \mathbb{R}$ is a continuous function and $\displaystyle \lim \limits_{x \to +\infty} f(x) = 1$. Find the following limit:
$$\large\displaystyle \lim \limits_{n \to \infty} \int \limits^{2006}_{1385}f(nx)\, \mathrm dx$$
 A: Given $\epsilon > 0$, there is some $M$ for which $|f(x) - 1| \leq \epsilon$ for $x > M$.
Then you can show that
$\left| \int_{1385}^{2006} f(nx) dx - (2006-1385)\right| \leq (2006-1385)\epsilon$ 
for sufficiently large $n$ (when plugging the left bound of the integrand, $1385n > M$). Finally take $\epsilon \to 0$.
A: Using the substitution $t=nx$, we get $I_n =  \int^{2006}_{1385}f(nx)dx =  \frac{1}{n}\int_{1385n}^{2006n} f(t) dt$. Let $I=2006-1385$.
Now let $\epsilon>0$, and choose $L>0$ such that if $t\ge L$, then $-\frac{\epsilon}{I} < f(t)-1 < \frac{\epsilon}{I}$.
Now choose $N\ge \frac{L}{1385}$. Then if $n \ge N$ and $t \in [1385n,2006n]$, we have $-\frac{\epsilon}{I} < f(t)-1 < \frac{\epsilon}{I}$. Integrating over $[1385n,2006n]$ and dividing by $n$ gives
$$ -\epsilon  < I_n -I < \epsilon $$
It follows that $\lim_n I_n = I$.
A: From continuity there is a $\xi_ n \in (1385,2006)$ such that $\int _{1385}^{2006}f(n x) \mathrm{d}x = (2006-1385)f(n \xi _n)$ . 
Since $k_n= n \xi _n \to \infty$ we have
\begin{align}\lim _{n\to \infty}\int _{1385}^{2006}f(n x) \mathrm{d}x &=\lim _{n \to \infty} (2006-1385)f(n \xi _n)\\
&=(2006-1385)\lim _{n \to \infty} f(k_n)=(2006-1385)\end{align}
A: Let $n\in \Bbb N$ and define $\displaystyle I_n:=\int \limits _{1385}^{2006}f(nx)\,\mathrm dx-621$.
It holds that  $\displaystyle I_n =\int \limits _{1385}^{2006}f(nx)-1 \,\mathrm dx$, therefore $\displaystyle |I_n|\leq \int \limits _{1385}^{2006} \left \vert f(nx)-1\right \vert \,\mathrm dx$.
Recall $\lim \limits_{x\to +\infty}f(x)=1 \iff (\forall \delta >0)(\exists \varepsilon >0)(\forall \overline x\in \Bbb R)(\overline x>\varepsilon \implies |f(\overline x)-1|<\delta)$.
Take $\delta >0$. There exists $\varepsilon >0$ such that for all $\overline x>\varepsilon$ it holds that $|f(\overline x)-1|<\delta$. In particular, for large enough $n$, if $x\ge 1385$, then $|f(nx)-1|<\delta$.
It follows that $\displaystyle |I_n|\leq \int \limits _{1385}^{2006}\delta \,\mathrm dx=621\delta$ and $\lim \limits _{n\to +\infty}\left( |I_n|\right)\leq \lim \limits _{n\to +\infty}\left(621\delta\right)=621\delta$.
Since $\delta$ was an arbitrary positive real number, it was proved that $\lim \limits _{n\to +\infty}\left( |I_n|\right)$ is a lower bound of $\{621\delta \colon\delta >0\}$, therefore $\lim \limits _{n\to +\infty}\left( |I_n|\right)\leq \inf \left(\{621\delta \colon \delta >0\} \right)=0$.
Finally $$\displaystyle 0=\lim \limits _{n\to +\infty}\left( |I_n|\right)=\lim \limits _{n\to +\infty}\left( I_n\right)=\lim \limits _{n\to +\infty}\left( \int \limits _{1385}^{2006}f(nx)\,\mathrm dx-621\right),$$ thus $\displaystyle \int \limits _{1385}^{2006}f(nx)\,\mathrm dx =621.$
A: Hint The Integral Operator is continuous, so you can swap limit and integral. If you have $f \searrow 1$ (monotonously).
