# Show that if $f$ is continuous on $[0, \infty)$ and $\lim_{x \to \infty} f(x) = a$ that $\lim_{x \to \infty} 1/x \int_0^x f(t)dt = a$.

Intuitively, I see this statement as saying that if a continuous function has a finite limit as x tends to infinity, the average value of the function is that same limit (which makes sense, since there are "infinitely many" contributions of that value to the average, very roughly speaking). I'm having trouble proving it with any rigour though. It looks like the fundamental theorem of calculus could be applied at first glance, but there really doesn't seem to be any use for it.

• Delta epsilon limit definition handles this quickly – davidlowryduda Feb 24 '14 at 10:44
• L'Hôpital will give the result immediately. – David Mitra Feb 24 '14 at 10:45
• If the integral is convergent it is not good idea to use L'Hospital rule. – kmitov Feb 24 '14 at 10:46
• The integral necessarily does not converge if it has a finite limit as x goes to infinity, unless a = 0 but then the result is trivial. – Andrew Martin Feb 24 '14 at 10:48
• @kmitov You could still use it (though it would be a bit much). The rule applies if the denominator has infinite limit (it doesn't matter what the numerator does). In any case, the method of the answer below is more edifying. – David Mitra Feb 24 '14 at 10:49

Suppose that $f(x) \to 0$. Then for any $\varepsilon>0$ there exists $A>0$ fixed such that $|f(x)|<\varepsilon$ for all $x>A$.
Therefore $\frac{1}{x}|\int_0^xf(t)dt| \le \frac{1}{x}\int_0^A|f(t)|dt + \frac{1}{x}\int_A^x|f(t)|dt \le \frac{A}{x}\sup_{t \in [0,A]}|f(t)|+\varepsilon\frac{x-A}{x}$ The first term can be made $<\varepsilon$ the second too.
If the limit is not zero you can write $\frac{1}{x}\int_0^x(f(t)-a)dt+\frac{1}{x}\int_0^x a dt$
• Thanks @kmitov. I believe you mean $|f(x)| < \varepsilon$ on the first line. – Andrew Martin Feb 24 '14 at 10:54
• I'm having some trouble showing that the first term can be $< \varepsilon$...we don't know anything about $f(t)$ in the interval other than the fact that it is continuous, and having $x > A$ isn't helping much. Am I missing something? – Andrew Martin Feb 24 '14 at 12:05
• If $f(t)$ is continuous on $[0,A]$ it is bounded. That is there exists $M>0$ such that $|f(t)|<M$ for all $t \in [0,A]$. Therefore, $\frac{A}{x}\sup_{t \in [0,A]}|f(t)| < \frac{A}{x}M.$ Since, $x \to \infty$ this can be made infinitely small. – kmitov Feb 24 '14 at 13:23
• I first thought that as well, but this needs to be true for all x > A, not some x. EDIT: Never mind...I just realized we can pick $K \geq A$ satisfying this, and we still have the limit definition holding. Thanks again for the help. – Andrew Martin Feb 24 '14 at 13:32