# 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.

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.

• L'Hospital's Rule works when denominator tends to $\infty$. There is no need to worry about numerator in this case. The problem here is that the limit after L'Hospital's Rule may not exist. – Paramanand Singh Sep 18 at 13:54
• No, it's a critical hypothesis. Once can a function doesn't goes to $\infty$, and real limit was $0$, but its drivation goes to $\infty$ and make wrong results. For example suppose $f(x)=\sqrt{1-x^2}$ on $[-1,1]$ that repeated in every $[2n-1,2n+1]$. – Ali Ashja' Sep 18 at 14:15
• You should check the proof for L'Hospital's Rule in this case rather than trying to find examples. A good reference is Wikipedia : en.wikipedia.org/wiki/L'H%C3%B4pital's_rule – Paramanand Singh Sep 18 at 14:17
• In your reference also it's mentioned that BOTH of them must goes to $\infty$. – Ali Ashja' Sep 18 at 14:23
• Wikipedia mentions this explicitly. "In case 2 the assumption that $f(x)$ diverges to infinity was not used in the proof..." – Paramanand Singh Sep 18 at 14:27

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?

• I didn't say that $g(t)=f(c)$. From the mean value theorem, I concluded that $g(t)=tf(c)$. So, then $g\prime(t)=f(c)$ – tomriddle99 Sep 18 at 5:26
• Even that is not true! For example, let $f(x) = x$ on $[0,1]$ and extended to a periodic continuous function on $\mathbb R$. We have $g(1) = \int_0^1 x dx = \frac 12 =f(\frac 12)$, but then it is not true that $g'(1) = f(\frac 12)$, in fact it equals $1 = f(1)$ by the fundamental theorem of calculus. – астон вілла олоф мэллбэрг Sep 18 at 5:32
• Okay. I got it. Thanks. – tomriddle99 Sep 18 at 5:36
• You are welcome! – астон вілла олоф мэллбэрг Sep 18 at 5:49