Finding a certain limit Let $f: [0,\infty)\to \mathbb{R}$ such that 
$$\lim_{x\to\infty}f(x)=m,$$
for some $m\in \mathbb{R}$. Let $\ell>0$ fixed. Find $\lim_{x\to\infty }\int_x^{x+\ell}f(s)ds$.
I think the answer is $m\ell$. I don't know if it's correct to say that as $x\to \infty$, $f(x)$ becomes contentious and monotone function. (as $f(x)$ gets closer to $m$.) Hence the above limit should be $m\ell$. I need some hints to be able to be more precise. Thank you.  
 A: Consider $\left|\displaystyle{\int_x^{x+\ell} f(x)\,dx} - m\ell\right| = \left|\displaystyle{\int_x^{x+\ell} \big(f(x)-m\big)\,dx}\right| \le \displaystyle{\int_x^{x+\ell} \left|f(x)-m\right|dx}$. Then use the definition of $\lim\limits_{x\to\infty} f(x)=m$.  (But, no, you cannot assume that $f$ is either continuous or monotone as you approach infinity. On the other hand, you don't need to assume such things! As someone else pointed out, you do need to assume $f$ is integrable on finite intervals so that the problem makes sense.)
A: Correct me if wrong:
$\lim_{x\rightarrow \infty} f(x) = m$ , assume $m \gt 0$ for definiteness.
For $\epsilon \gt 0$ there is an $x_0$  $\in \mathbb{R^+}$ ,such that 
$| f(x) - m | \lt \epsilon$ for $x \gt x_0$:
$m - \epsilon \lt f(x) \lt m + \epsilon$.
Assume $f$ to be integrable in the range considered,  integrate from $x$ to $x + l$, monotony implies:
$(m - \epsilon)l  \lt  \int_{x}^{x+l} f(x) dx  \lt (m + \epsilon)l$.
Since $\epsilon$ is arbitrary: 
$\lim_{x \rightarrow \infty} \int_{x}^{x + l} f(x) dx = ml$.
To recap more formally:
For every $\epsilon$ (set $\epsilon /l$ in the above derivation) there exists a  $x_0 \in \mathbb{R^+}$ such that  $x \gt x_0$ implies 
$|\int_{x}^{x +l} f(x) dx - ml| \lt \epsilon$, I.e.
$\lim_ {x \rightarrow \infty} \int_{x}^{x+l} f(x) dx = ml$.
