if $f$ is continous and $ \lim_{x\to \infty} f=a$ prove that $\lim\limits_{x \to \infty} \frac{1}{x} \int_0^x f=a$ So Spivak has done a lot of work and I've done something too simple to be correct, so lets my answers:
since:
 $ \lim_{x\to \infty} f=a$
There exists an $\delta$ such that $x>\delta$ $\to$ $|f-a|<\varepsilon$
so $-\varepsilon+a<f<a+\varepsilon.$ Since $f$ is continuous it attains its minimum and maximum value on some argument, so the Lower sum of Darboux $L(F,P)$ and Upper sum $U(F,P)$,where P is a partition of $[0,x]$, have the following propertie:
$\int_0^x-\varepsilon+a<L(F,P)\le \int_0^x f \leq U(F,P) \leq \int_0^x\varepsilon+a$
So $-\varepsilon+a \lt \frac{\int_0^x f}{x}\lt \varepsilon+a$
Concluding: given $\varepsilon>0$ , $\exists$ $x>\delta$ such that $|\frac{\int_0^x f}{x}-a|$ 
So is this right?
 A: $g(x):=|\frac{\int_0^xf(t)dt}{x}-a| \leq \frac{\int_0^x|f(x)-a|dx}{x}$
Let $\epsilon>0$
Exists $M>0$ such that $|f(x)-a| < \epsilon, \forall x \in (M,+\infty)$
Let $x>M$
$$ \frac{\int_0^x|f(x)-a|dx}{x}=\frac{\int_0^M |f(x)-a|dx}{x}+\frac{\int_M^x |f(x)-a|dx}{x}$$ $$\leq \frac{\int_0^M |f(x)-a|dx}{x}+ \epsilon (\frac{x-M}{x})$$
The function $|f(x)-a|$ is continuous on $[0,M]$ thus bounded
so $\frac{\int_0^M |f(x)-a|dx}{x} \leq \frac{M(C+|a|)}{x}$
So $\limsup_{x \to +\infty}g(x) \leq \epsilon$
Since $\epsilon>0$ arbitrary we have that $\limsup_{x \to +\infty}g(x)=0$
So $\lim_{x \to  +\infty}g(x)=0$
A: Let $\epsilon>0. $Choose $y$ so large that $|f(c)-a|<\epsilon$ whenever $c>y.$ The mean value theorem, the fundamental theorem of calculus and the sum property of the integral imply that
$\frac{\int^x_0f(t)dt}{x}=\frac{\int^y_0f(t)dt}{x}+\frac{\int^x_yf(t)dt}{x}=\frac{\int^y_0f(t)dt}{x}+\frac{f(c)(x-y)}{x}$ for some $y<c<x$. Let $x\to \infty.$ 
Then, $\frac{\int^x_0f(t)dt}{x}\to f(c).$ But, we have arranged it so that $|f(c)-a|<\epsilon$ so 
$\left |\underset{x\to \infty}\lim\frac{\int^x_0f(t)dt}{x}-a\right|<\epsilon.$ Since $\epsilon$ is arbitrary, the result follows. 
A: Note, that the case $|a| \neq 0$ follows directly with L'Hospital rule:


*

*$a>0$: Choose $M>0$ such that $f(x) > \frac{a}{2}$ for $x>M$
$$\stackrel{x>M}{\Rightarrow} \int_0^xf\,dt= \underbrace{\int_0^Mf\,dt}_{K:=} + \int_M^xf\,dt\geq K+\underbrace{(x-M)\frac{a}{2}}_{\stackrel{x\to \infty}{\longrightarrow}\infty}\stackrel{x\to \infty}{\longrightarrow}\infty$$
$$\Rightarrow\frac{1}{x}\int_0^xf\,dt \stackrel{L'Hosp.}{\sim}\frac{f(x)}{1}\stackrel{x\to \infty}{\longrightarrow}a$$

*$a<0$: Just consider $-f$.

*$a=0$: For $\epsilon >0$ choose $M_{\epsilon}$ such that $|f(x)| < \frac{\epsilon}{2}$ for $x > M_{\epsilon}$ and set $K_{\epsilon} = \int_0^{M_{\epsilon}}|f|\,dt$
$$\stackrel{x>\max(M_{\epsilon},\frac{2K_{\epsilon}}{\epsilon})}{\Longrightarrow}\left| \frac{1}{x}\int_0^xf\,dt\right| \leq  \underbrace{\frac{K_{\epsilon}}{x}}_{< \frac{\epsilon}{2}} + \underbrace{\frac{(x-M_{\epsilon})\epsilon}{2x}}_{<\frac{\epsilon}{2}}< \epsilon$$
