Prove $\lim\limits_{x\to\infty}\frac{1}{x}\int_0^xf(t)dt=a$ Let $f:[0,\infty)\to\mathbb{R}$ continuous and $\lim\limits_{x\to\infty}f(x)=a$. Claim: $\lim\limits_{x\to\infty}\frac{1}{x}\int_0^xf(t)dt=a$.

My try: It is $\int_0^xf(t)dt=F(x)-F(0)$ (because of the fundamentaltheorem of calculus) and $\lim\limits_{x\to\infty}F(x)=ax+b$, because $\lim\limits_{x\to\infty}f(x)=a$.
What can I do next, or do you have an idea how to prove it? Regards
Use L'Hôpital's rule and the fundamental theorem of calculus. There are two cases.

Case 1: $\int_0^\infty f(t)dt = \pm \infty$. In this case, by L'Hôpital's rule and the FTC, $$\lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)dt = \lim_{x \to \infty}\frac{\frac{d}{dx}\int_0^x f(t)dt}{\frac{d}{dx}x} = \lim_{x \to \infty}\frac{f(x)}{1} = a \text{.}$$

Case 2: $\int_0^\infty f(t)dt$ is finite. Then the limit in question is clearly 0, and we must show that $a = 0$ as well. However, if $a \neq 0$, then $f$ is eventually bounded away from 0, and so the improper integral would be infinite (a contradiction).
You're close: as $\;f\;$ is continuous it has a primitive $\;F\;$ , so

$$\frac1x\int_0^xf(t)dt=\frac{F(x)-F(0)}{x-0}\stackrel{MVT}=F'(c_x)=f(c_x)\;,\;\;c_x\in(0,x)$$

and now just take the limit $\;x\to\infty\;$ ... and be careful (for example, for different $\;x$'s one could possibly get the same $\;c_x\;$ . How do you go around this?)