I have a function $F(x) = \frac{1}{x} \int_0^x f(t) \, dt$. If $f(x) \to L$ show that $F(x) \to L$ as $x\to\infty$.
So far i have tried to split the integral up like so $$ \lim_{x\to\infty}\frac{1}{x}\left(\int_0^bf(t) \, dt + \int_b^x f(t) \, dt \right) $$ but a bit further along this I get stuck
Given $\varepsilon>0$, we want to show that whenever $x$ is bigger than some number $x_0$ (which depends on $\varepsilon$) the difference between $L$ and $\frac 1 x\int_0^x f(t)\,dt$ is less (in absolute value) than $\varepsilon$.
We know that there exists $x_1$ such that whenever $t>x_1$ then $|f(t)-L|<\varepsilon/2$.
So look at the difference \begin{align} \left|\frac 1 x \int_0^x f(t)\,dt - L\right| & = \frac 1 x \left|\int_0^x (f(t)-L)\,dt \right| \\[10pt] & = \frac 1 x \int_0^{x_1} (f(t)-L)\,dt + \frac 1 x \int_{x_1}^x (f(t)-L)\,dt. \end{align} The first integral is $1/x$ times something not changing as $x$ changes. Hence it approaches $0$ as $x\to\infty$. It can be made less in absolute value than $\varepsilon/2$ by making $x$ big enough. For the second integral we have $$ \left|\frac 1 x \int_{x_1}^x (f(t)-L)\,dt \right| \le \frac 1 x \int_{x_1}^x |f(t)-L|\,dt\le\frac 1 x\int_{x_1}^x \frac\varepsilon2\,dx \le \frac{x-x_1} x \frac\varepsilon2 \le\frac\varepsilon2. $$