Find the condition on $t$ if the limit is defined. $f$ is a real and continuous function defined by $f(x+1)=f(x) \forall x\in\mathbb{R}$. $g(t)=\int_0^tf(x)\,dx$, $t \in \mathbb{R}$. Then $h(t)=$$\lim\limits_{n\to\infty}\frac{g(t+n)}{n}$ is defined for :
(a) $t=0$ only
(b) integer $t$ only
(c) $\forall t \in \mathbb{R}$ and $h(t)$ is independent of $t$
(d) none of the above is true
My attempt:
$g(t)= \int_0^tf(x)dx=f(c)(t-0) $ for some $c\in(0, t)$ by mean value theorem. $\implies g\prime(t)=f(c)$.Now, $h(t) = \lim\limits_{n\to\infty}\frac{g(t+n)}{n} = \lim\limits_{n\to\infty}\frac{g\prime(t+n)}{1}$. Now, as $g\prime=f $ is periodic, we can say, $h(t)=\lim\limits_{n\to\infty} g\prime(t)=f(c)$. Therefore option (c) is the correct answer. But the problem with the proof is that nowhere it is said that $n$ is a natural number. But I am using the periodic property taking $n$ as a natural number. So, how can I correct this proof? Should I ignore the non-integer part as $\lim\limits_{n\to\infty}\frac{\{n\}}{n}$ goes $0$? Thanks. 
 A: At first, you make $2$ mistakes:
First one: The number $f(c)$, in fact, is dependent to variable $t$, that may changes by different $t$, so the true derivation:
$$g'(t) = t\frac{d}{dt}f(c) + f(c)$$
Second one: You use L'Hôpital's Rule when your denominator goes to $\infty$, so it was allowed iff the nominator also goes to $\infty$, that was not true, for example let $f(x)=0$.
But you can do as follow:
$$h(t) = \lim_{n \longmapsto \infty} \frac{g(t+n)}{n} = \lim_{n \longmapsto \infty} \frac{\int_0^{t+n}f(x)dx}{n} = \lim_{n \longmapsto \infty} \frac{\int_0^nf(x)dx + \int_n^{t+n}f(x)dx}{n} =$$
$$\lim_{n \longmapsto \infty} \frac{n \int_0^1f(x)dx + \int_0^tf(x)dx}{n} = \lim_{n \longmapsto \infty} (\int_0^1f(x)dx + \frac{g(t)}{n}) = \int_0^1f(x)dx + \lim_{n \longmapsto \infty} \frac{g(t)}{n}$$
Now as $f$ is a continuous periodic function, it's bounded, and so for every $t$, $g(t)$ is finite, so we have:
$$h(t) = g(1)$$
So the option c is correct, but in this way you can prove it.
A: From $g(t) = tf(c)$ for some $c \in (0,t)$, it does $\color{red}{\text{not}}$ follow that $g'(t) = f(c)$ for that same $c$. In fact, by the fundamental theorem of calculus, we have that $g'(t) = f(t)$ for all $t$. So that statement is wrong.
Everything that follows from here is then flawed by the above mistake.

What you need to do is notice that for any positive integer $n$ we have $g(n) = \underbrace{g(1)+g(1)+... + g(1)}_{n \text{ times}} = ng(1)$, by periodicity of $f$ (draw a diagram of any $f$ and see this for yourself).
In a similar fashion, for any $t \in \mathbb R$ and positive integer $n$, we have $g(t+n) - g(t) = ng(1)$, by noting that $g(t+n) - g(t) = \int_t^{t+n} f(x)dx$ and then using periodicity.
Therefore, for any $t \in \mathbb R$ and $n$ a positive integer, we have :
$$
\frac{g(t+n)}{n} = \frac{g(t+n) - g(t)}{n} + \frac{g(t)}{n} = g(1) + \frac{g(t)}{n}
$$
Now, what happens as $n \to \infty$? Can you conclude?
