All functions such that $\lim_{n\to \infty} \int_n^{n+\frac 1n} f(x)dx$ is a non-zero finite value I’m interested in finding all functions $f(x)$ (if any), such that 
$$\lim_{n\to \infty} \int_n^{n+\frac 1n} f(x)dx$$ is a finite non-zero value. 
Let’s try evaluating this limit for a few functions that one would expect to follow this property. All I could deduce was that $\lim_{x\to \infty} f’(x) \ne 0$ otherwise our limit would be zero.
$$\int_n^{n+\frac 1n} x^2 \ dx = \frac13 \left[ \left(n + \frac 1n \right)^3 - n^3\right] \\ = \frac{3n^2 + 3+ \frac{1}{n^2}}{3n}$$
It’s easy to see that the limit of this diverges to $\infty$.
$$\int_n^{n+\frac 1n} \sqrt x \ dx = \frac 23 \cdot \frac{3n + \frac 3n + \frac{1}{n^3}}{\left(n + \frac 1n \right)^{\frac 32} + n^{\frac 32} } $$
This time, the limit goes to $0$. So, there must be one (or more than one?) value of $k$ with $\frac 12 \lt k \lt 2$ such that $f(x)=x^k$ satisfies our property. I saw that $\lim_{n\to \infty} \int_n^{n+\frac 1n} x \ dx = 1$
Are there other such functions? Can this be generalized?
 A: If we assume that $f$ is continuous we have, by the mean value theorem, that $$\int_{n}^{n+1/n}f\left(x\right)dx=\frac{f\left(c_{n}\right)}{n}$$ where $c_{n}\in\left[n,n+1/n\right]$. Hence $$\lim_{n\rightarrow+\infty}\frac{f\left(c_{n}\right)}{n}=\lim_{n\rightarrow+\infty}\frac{f\left(n\right)}{n}$$ and this limit is finite if and only if $f\left(x\right)\sim ax,\,a\in\mathbb{R}$, as $x\rightarrow+\infty$.
A: We want to consider the integral $$I=\int\limits ^{n+1/n}_{n} f( x) dx$$
Where $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous.
Suppose $f$ has antiderivative $F$. Then $$I=F(n+\frac{1}{n})-F(n)$$
Using a first order Taylor approximation, $$F\left( n+\frac{1}{n}\right) \approx F( n) +F'( n)\left(\frac{1}{n}\right)$$
And since $F'=f$, $$I \approx \frac{f(n)}{n}$$
And this approximation will clearly get better and better for larger and larger $n$. Therefore for $\lim _{n\rightarrow \infty }\frac{f( n)}{n}$ to be finite and nonzero, $\text{deg}(f)=1$. So linear equations are the only types of equations that have this property. 
EDIT: functions with linear-like growth will also have this property. So functions such as $\frac{x}{\log(x)}$, $x+e^{-x}$, and so on.
A: If $f(x)$ is a polynomial each term is like $ax^m$. The definite integral is 
$\frac{a}{m+1}(n+\frac{1}{n})^{m+1}-\frac{a}{m+1}n^{m+1}$
$=\frac{a}{m+1}(n^{m+1}+(m+1)n^m(\frac{1}{n})^{1}+\frac{(m+1)m}{2}n^{m-1}(\frac{1}{n})^{2}+ ...-n^{m+1})$
$=\frac{a}{m+1}((m+1)n^m(\frac{1}{n})^{1}+\frac{(m+1)m}{2}n^{m-1}(\frac{1}{n})^{2}+ ...)$
$=an^m(\frac{1}{n})+a\frac{m}{2}n^{m-1}(\frac{1}{n})^{2}+ ...)$
$=an^{m-1}+a\frac{m}{2}n^{m-3}+ ...$
If $m>1$ the first term becomes infinite as $n\to\infty$. If $m<1$ the first term (and later terms) approach zero as $n\to\infty$. If $m=1$ then the first term is $an^0=a$ and later terms approach zero. So the only polynomial function is $f(x)=ax$
