Problems with calculating the limit $\lim_{x\to \infty} \frac{1}{x}\int_0^x{\lvert \sin(t)\rvert dt}$

Question

I wanted to check whether theorem below is true, for some simple case: $$\lvert\lim_{x\to \infty}f(x)\rvert=\infty \Rightarrow \lim_{x\to \infty} \frac{1}{x}\int_0^x{\lvert \sin(f(t))\rvert\, dt}=\frac{2}{\pi}$$ And I let $f(x)=x$ $$\lim_{x\to \infty} \frac{1}{x}\int_0^x{\lvert \sin(f(t))\rvert\, dt}=\lim_{x\to \infty} \frac{\int_0^x{\lvert \sin(t)\rvert\, dt}}{x}$$ To calculate this limit we should be able to use L'Hospital's rule. $$\lim_{x\to \infty} \frac{\int_0^x{\lvert \sin(t)\rvert \,dt}}{x}=\lim_{x\to \infty} \frac{d}{dx}\int_0^x{\lvert \sin(t)\rvert\, dt}=\lim_{x\to \infty} \lvert \sin(x)\rvert$$ But this limit does not exist.
However I managed to calculate this limit in other way, firstly evaluating the integral. $$\int_0^x{\lvert \sin(t)\rvert\, dt}=\int_0^{\{\frac{x}{\pi}\}\pi}\lvert\sin(t)\rvert\,dt+\int_0^{\lfloor \frac{x}{\pi} \rfloor\pi}\lvert \sin(t)\rvert\, dt=\int_0^{\{\frac{x}{\pi}\}\pi} \sin(t)\, dt+\Bigl\lfloor\frac{x}{\pi}\Bigr\rfloor\int_0^{ {\pi} } \sin(t)\,dt= $$ $$=1-\cos(\{\frac{x}{\pi}\}\pi)+2\lfloor\frac{x}{\pi} \rfloor,$$ where $\{x\}$ denotes the fractional part of $x$
Applying this formula to the limit we get: $$\lim_{x\to \infty} \frac{\int_0^x{\lvert \sin(t)\rvert\, dt}}{x}=\lim_{x\to \infty}\frac{1-\cos(\{\frac{x}{\pi}\}\pi)+2\bigl\lfloor\frac{x}{\pi} \bigr\rfloor}{x}=\lim_{x\to \infty}\frac{2\bigl\lfloor\frac{x}{\pi} \bigr\rfloor}{x}=2 \lim_{x\to \infty}\frac{1}{x}\Bigl(\frac{x}{\pi}-\Bigl\{\frac{x}{\pi}\Bigr\}\Bigr)=\frac{2}{\pi}-\lim_{x \to\infty}{\frac{1}{x}\Bigl\{\frac{x}{\pi}\Bigr\}}=\frac{2}{\pi}$$ Which is supposed result. It seems that the second result is proper, but then the question is: why can't we apply L'Hospital's rule to this limit?
Thanks for all the help.

Answer 1

In L'Hospital's rule, $$ \lim_{x\to c}\frac{f'(x)}{g'(x)} $$ is supposed to exist. But in your problem, it doesn't exist. So you can't use this rule for this problem.