TITLE: Limit $\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$
QUESTION [7 upvotes]: Determine whether or not the limit below exists.
$$\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$$
I tried to use the Mean value theorem integrals to prove the limit exists, but it does not exist for $\lim\limits_{x\rightarrow0}\sin \frac{1}{x}$. So I guessed the limit does not exist and used the Cauchy principle to prove it, but I failed.
Any idea will be helpful.

REPLY [3 votes]: Well, here is a slightly different approach.
Note that $(\sin x) /x\to 1$ as $x\to 0$ and hence (add missing steps here if you need) the desired limit equals the limit of $$\frac {1}{x}\int_{0}^{x}\sin(1/t)\cos t^2\,dt$$ and we can write the above expression as $$\frac{1}{x}\int_{0}^{x}\sin(1/t)(\cos t^2-1)\,dt+\frac{1}{x}\int_0^x\sin(1/t)\,dt$$ The integrand in first integral tends to $0$ as $t\to 0$. Therefore by the fundamental theorem of calculus (part 1) the first term above tends to $0$.
The desired limit thus equals $$\lim_{x\to 0}\frac {1}{x}\int_0^x\sin(1/t)\,dt$$ This is a famous question on this website and the desired limit is $0$. You can prove this by using the auxiliary function $$g(x) =x^2\cos(1/x),x\neq 0,g(0)=0$$ We have $$g'(x) =2x\cos(1/x)+\sin(1/x),x\neq 0,g'(0)=0$$ By fundamental theorem of calculus (part 2) we have $$g(x) =g(x) - g(0)=\int_0^x g'(t) \, dt$$ which means that $$\frac {1}{x}\int_{0}^{x}\sin(1/t)\,dt=\frac{g(x)}{x}-2\cdot\frac{1}{x}\int_0^x t\cos(1/t)\,dt$$ Now the first term on right tends to $g'(0)=0$ and second term on right also tends to $0$ (via fundamental theorem of calculus (part 1) as the integrand tends to $0$ as $t\to 0$).
You should also observe that the above derivation works in case $\cos t^2$ is replaced by any Riemann integrable function $\phi(t)$ which tends to $1$ as $t\to 0$.

For those wondering about the version of FTC used in above I state them explicitly.
FTC Part 1: Let $f:[a, b] \to\mathbb {R} $ be a function which is Riemann integrable on $[a, b]$. Then the function $F:[a, b] \to\mathbb {R} $ defined by $$F(x) =\int_a^x f(t) \, dt$$ is continuous on $[a, b] $. Further if $c\in[a, b] $ is a point such that right hand limit of $f$ at $c$ is $L$ then the right hand derivative of $F$ at $c$ exists and equals $L$ (the statement holds if right is replaced by left).
FTC Part 2: Let $f:[a, b] \to\mathbb {R} $ be Riemann integrable on $[a, b] $ and let $F:[a, b] \to \mathbb {R} $ be continuous on $[a, b] $ and differentiable on $(a, b) $ with $F'(x) =f(x)\, \forall x\in(a, b) $ then $$\int_a^b f(x) \, dx=F(b) - F(a) $$