Hints.
- For the first one, you can take
$$x_n=\frac1{\sqrt{n+1}}.$$
You have
$$\lim_{n\to\infty }x_n=0$$
but
$$\lim_{n\to\infty} \cos\left(\frac 1{{x_n}^2}\right)=\lim_{n\to\infty} \cos(n)$$
which is undefined (why?).
- For the second one, you can take
$$y_n=e^n.$$
You have
$$\lim_{n\to\infty }y_n=+\infty$$
but
$${y_n}^{1+\sin(y_n)}=e^{(1+\sin(y_n))\ln(y_n)}=e^{(1+\sin(e^n))n}.$$
And the sequence $((1+\sin(e^n))n)_n$ comes infinitely many times as close to $0$ as you want (why?), and get infinitely many times as big as you want (why?).
$f'(x)=\frac{\text{d} f(x)}{\text{d}x}=\int_{0}^{\infty} \frac{\partial }{\partial x}\left(\frac{e^{-u}-e^{-u^{\alpha}x}}{u}\right) du$
Taking, $z=u^{\alpha}x$ we get
$=\frac{x^{-1}}{\alpha}(\int_{0}^{\infty} e^{-z} dz) =\frac{x^{-1}}{\alpha}$.....(1)
Taking, $f(x)=y$
So, from (1) we get the differential equation
$\frac{\text{d} y}{\text{d}x}=\frac{x^{-1}}{\alpha}$ with the boundary condition $y(1)=\int_{0}^{\infty} \frac{e^{-u}-e^{-u^{\alpha}}}{u} du =-\frac{(\alpha-1)}{\alpha}\gamma$
Proof: Take $y_{alpha}(1)=\psi(\alpha)=\int_{0}^{\infty} \frac{e^{-u}-e^{-u^{\alpha}}}{u} du$
So, $\frac{\text{d}\psi(\alpha)}{\text{d}x}=\int_{0}^{\infty} \frac{\partial}{\partial \alpha}\left(\frac{e^{-u}-e^{-u^{\alpha}}}{u}\right) du$
$=\int_{0}^{\infty} u^{\alpha-1}\text{ln}(u)e^{-u^{\alpha}} du$
Taking $u^{\alpha}=z$
$=\frac{1}{{\alpha}^2}\int_{0}^{\infty} \text{ln}(z)e^{-z} dz=-\frac{\gamma}{{\alpha}^2}$
As $\Gamma'(1)=-\gamma$ (https://en.m.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant)
So, we get the differential equation
$\frac{\text{d}\psi}{\text{d}\alpha}=\frac{-\gamma}{{\alpha}^2}$ with the boundary condition $\psi(1)=0$ we get
$y(1)=\psi(\alpha)-\psi(1)=\psi(\alpha)=-\frac{\alpha-1}{\alpha}\gamma$
And for, all $\alpha\geq 1$, $y(1)=\frac{-(\alpha-1)\gamma}{\alpha}$.
Where, $\gamma$ is Euler- Mascheroni constant.
So, $f(x)=\frac{\text{ln}x}{\alpha} -\frac{(\alpha-1)\gamma}{\alpha}$.
So, for $\alpha=1$, $f(x)=\text{ln}x$.
Best Answer
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.