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

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.
