Solution verification for evaluating $\lim\limits_{x \to +\infty}\dfrac{\int_0^{x}|\sin t|{\rm d}t}{x}$

Question

My Solution

Since $x \to +\infty$，we may assume $x>\pi$. Then $$ \exists n \in \mathbb{N}:n\pi\leq x< (n+1)\pi.$$ Thus $$2n=\int_0^{n\pi}|\sin t|{\rm d}t\leq \int_0^x|\sin t|{\rm d}t<\int_0^{(n+1)\pi}|\sin t|{\rm d}t=2(n+1).$$ Further $$\frac{2n}{(n+1)\pi}<\dfrac{\int_0^{x}|\sin t|{\rm d}t}{x}<\frac{2(n+1)}{n\pi}.$$ When $x \to +\infty$，$n \to \infty$. We may obatin $$\frac{2n}{(n+1)\pi},\frac{2(n+1)}{n\pi} \to \frac{2}{\pi}(n \to \infty).$$ By the squeeze theorem, it follows that $$\lim_{x \to +\infty}\dfrac{\int_0^{x}|\sin t|{\rm d}t}{x}=\frac{2}{\pi}.$$

Answer 1

As the function is periodic, the limit is also the average value over a single period (you sum arbitrarily many whole periods plus a single incomplete one, which is bounded), hence

$$\frac1\pi\int_0^\pi|\sin x|\,dx=\frac2\pi.$$

Answer 2

Yes, your solution is correct.

(Answering as community wiki so this question can be closed as answered.)