Limit of an integral of a continuous function Let $f$ a continuous function in $\mathbb{R}_+$ and suppose that $\lim_{x \to \infty}f(x)=a$. Show that: $$\lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)dt = a$$
I considered the Fundamental Theorem of Calculus, getting that above limit is equal to: $$\lim_{x \to \infty} \frac{F(x)-F(0)}{x}$$
However, I'm not sure how to conclude the equality with $a$ from this step. 
 A: Let $\varepsilon >0$ and choose $R\in \mathbb{R}$ such that 
$$|f(x)-a| < \varepsilon $$ if $x>R$. Then, for any $s>R$ 
$$(s-R)(a-\varepsilon) < \int_R^s f(x) < (s-R)(a+\varepsilon)$$
Now divide by $s$ and let $s\rightarrow \infty$.
Edit: Note: just to make sure you do not overlook this: this quietly assumes that $\int_0^Rf(x)dx$ is finite for any finite $R>0$, which is true, for example, if $f$ is continuous on $[0,\infty)$. If $f$ is continuous only on $(0, \infty)$, this may not work (it is still sufficient if the integral from $0$ to $R$ is finite for any $R$, you just cannot prove this anymore) and the statement you are after might actually not be true.
A: First, it is NOT the definition of the derivative. (Better check it out in your textbook yourself)
For a short proof.  Your limit is just $\lim_{x\to+\infty}F(x)/x$.  Now that since the denominator goes to infinity, and $\lim_{x\to+\infty}F'(x)/x'=\lim_{x\to+\infty}f(x)=a$ exists. By l'Hospital we can claim that  $\lim_{x\to+\infty}F(x)/x=a$. 
(PS: in the strongest version of l'Hospital we don't necessarily need both the numerator and denominator tends to infinity simultaneously. We only require the latter)
A: Using L'Hospital's Rule, we have
$$\lim_{x\to \infty}\frac{\int_0^xf(t)dt}{x}=\lim_{x\to \infty}f(x)=f(a)$$
NOTE:
L'Hospital's Rule does not require that the limit of the numerator approach $\infty$ as shown in Note 2 of THIS REFERENCE.
