Does$$\lim_{h\to 0} \frac1h\int_0^h \sin(1/t)\,dt\tag1$$ exist?
Let $F(h)=\int_0^h \sin(1/t)\,dt$. There is no closed form expression for $F$. If the limit in $(1)$ existed, then $F(h)$ would be an example of a differentiable function whose derivative is discontinuous, which is why I am interested. Solutions which are comprehensible by an exceptionally bright high school calculus student would be preferred, if possible.
My thoughts:
It is tempting to apply l'Hopital's rule to $\lim_{h\to 0}\frac{F(h)}{h}$, but $\lim_{h\to 0}\frac{F'(h)}{1}=\lim_{h\to 0}\sin(1/h)$ does not exist.
An equivalent expression, obtained from substituting $1/u$ for $x$, is $$ \lim_{h\to 0}\frac1h\int_{1/h}^{\infty}\frac{-\sin u}{u^2}\,du=\lim_{a\to\infty}a\cdot\int_a^\infty \frac{-\sin u}{u^2}\,du.\tag2$$ It is tempting to bound $\left|\int_a^{\infty}-\sin u/u^2\,du\right|$ by $\int_a^{\infty}1/u^2\,du=1/a$, but this bound is too coarse.
I strongly suspect the limit exists and is equal to zero. In $(2)$, I think the alternating sign of $\sin u$ causes $\int_a^\infty \frac{-\sin u}{u^2}\,du$ to go to zero faster than $a$. However, I am getting tongue-tied in making this argument formal, especially using the tools available to a high-school calculus student (l'Hopital's, MVT, integration techniques, etc).
Thank you in advance for any help.
