Limit $\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$

Since $\cos t^2$ is monotone decreasing for $t$ sufficiently close to $0$, by the second mean value theorem for integrals there exists $\xi_x \in (0, \sin x)$ such that

$$\int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = \cos(0) \int_0^{\xi_x}\sin \frac{1}{t} \, dt = \int_0^{\xi_x}\sin \frac{1}{t} \, dt$$

Taking $g(t) = t^2 \cos \frac{1}{t}$ for $t > 0$ and $g(0) = 0$, we have $g’(0) =0$ and for $t>0$,

$$g'(t) = 2t \cos \frac{1}{t} + \sin \frac{1}{t},$$ and, using the FTC,

$$\int_0^{\xi_x}\sin \frac{1}{t} \, dt = \int_0^{\xi_x} g'(t) \, dt-\int_0^{\xi_x}2t \cos \frac{1}{t} \, dt=\xi_x^2\cos \frac{1}{\xi_x} - \int_0^{\xi_x}2t \cos \frac{1}{t} \, dt$$

Thus,

$$\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = \xi_x \frac{\xi_x}{x}\cos \frac{1}{\xi_x} - \frac{1}{x} \int_0^{\xi_x}2t \cos \frac{1}{t} \, dt$$

We can apply the mean value theorem to the integral on the RHS (since the integrand is continuous) to find $\theta_x \in (0,\xi_x)$ such that

$$\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = \xi_x \frac{\xi_x}{x}\cos \frac{1}{\xi_x} - \frac{\xi_x}{x} 2\theta_x \cos \frac{1}{\theta_x} $$

Since $\xi_x/x < 1$, we have

$$\left|\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt\right| \leqslant \xi_x +2 \theta_x$$

Since $\xi_x , \theta_x \to 0$ as $x \to 0+$, we get

$$\lim_{x \to 0+}\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = 0$$

Similarly we can show that the limit as $x \to 0-$ is $0$ as well.

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 $t\to 0$.

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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 $$\int_a^b f(x) \, dx=F(b) - F(a) $$