Show that
$$\int _ { 0 } ^ { 1 } \ln \left( \frac { 1 - e ^ { - t u } } { t } \right) \mathrm { d } u \underset { t \rightarrow 0 ^ { + } } { \longrightarrow } - 1$$
You may begin by establishing that $x \mapsto \frac { 1 - e ^ { - x } } { x }$ is decreasing on $\mathbf { R } _ { + } ^ { * }$.

Show that $$\int_{0}^{1} \ln\left(\frac{1-e^{-tu}}{t}\right) \mathrm{d}u \underset{t \rightarrow 0^+}{\longrightarrow} -1.$$ One may begin by establishing that $x \mapsto \frac{1-e^{-x}}{x}$ is decreasing on $\mathbf{R}_+$.

For each positive integer $n$, define a function $f_n$ on $[0,1]$ as follows: $$f _ { n } ( x ) = \left\{ \begin{array} { c c c } 0 & \text { if } & x = 0 \\ \sin \frac { \pi } { 2 n } & \text { if } & 0 < x \leq \frac { 1 } { n } \\ \sin \frac { 2 \pi } { 2 n } & \text { if } & \frac { 1 } { n } < x \leq \frac { 2 } { n } \\ \sin \frac { 3 \pi } { 2 n } & \text { if } & \frac { 2 } { n } < x \leq \frac { 3 } { n } \\ \vdots & \vdots & \vdots \\ \sin \frac { n \pi } { 2 n } & \text { if } & \frac { n - 1 } { n } < x \leq 1 . \end{array} \right.$$ Then, the value of $\lim _ { n \rightarrow \infty } \int _ { 0 } ^ { 1 } f _ { n } ( x ) d x$ is
(A) $\pi$
(B) 1
(C) $\frac{1}{\pi}$
(D) $\frac{2}{\pi}$

Let $$f ( x ) = e ^ { - | x | } , x \in \mathbb { R }$$ and $$g ( \theta ) = \int _ { - 1 } ^ { 1 } f \left( \frac { x } { \theta } \right) d x , \theta \neq 0$$ Then, $$\lim _ { \theta \rightarrow 0 } \frac { g ( \theta ) } { \theta }$$ (A) equals 0 .
(B) equals $+ \infty$.
(C) equals 2 .
(D) does not exist.

Let $\psi : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with $\int_{-1}^{1} \psi(x)\,\mathrm{d}x = 1$. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function. Then $$\lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} \int_{1-\varepsilon}^{1+\varepsilon} f(y)\,\psi\!\left(\frac{1-y}{\varepsilon}\right) \mathrm{d}y$$ equals
(A) $f(1)$
(B) $f(1)\psi(0)$
(C) $f'(1)\psi(0)$
(D) $f(1)\psi(1)$

For $x \in \mathbb { R }$, let $\tan ^ { - 1 } ( x ) \in \left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$. Then the minimum value of the function $f : \mathbb { R } \rightarrow \mathbb { R }$ defined by $f ( x ) = \int _ { 0 } ^ { x \tan ^ { - 1 } x } \frac { e ^ { ( t - \cos t ) } } { 1 + t ^ { 2023 } } d t$ is

Let $\alpha$ and $\beta$ be the real numbers such that
$$\lim _ { x \rightarrow 0 } \frac { 1 } { x ^ { 3 } } \left( \frac { \alpha } { 2 } \int _ { 0 } ^ { x } \frac { 1 } { 1 - t ^ { 2 } } d t + \beta x \cos x \right) = 2$$
Then the value of $\alpha + \beta$ is $\_\_\_\_$ .