We consider the function $f$ defined on $\mathbb { R }$ by:
$$f ( x ) = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { x } + 1 }$$
The representative curve $\mathscr { C }$ of the function $f$ in an orthonormal coordinate system is given.

1. Calculate the limit of the function $f$ at negative infinity and interpret the result graphically.
2. Show that the line with equation $y = 2$ is a horizontal asymptote to the curve $\mathscr { C }$.
3. Calculate $f ^ { \prime } ( x )$, where $f ^ { \prime }$ is the derivative function of $f$, and verify that for all real numbers $x$ we have: $$f ^ { \prime } ( x ) = \frac { f ( x ) } { \mathrm { e } ^ { x } + 1 } .$$
4. Show that the function $f$ is increasing on $\mathbb { R }$.
5. Show that the curve $\mathscr { C }$ passes through the point $\mathrm { I } ( 0 ; 1 )$ and that its tangent at this point has slope 0.5.

For a constant $a > 3$, two curves $y = a ^ { x - 1 }$ and $y = 3 ^ { x }$ meet at point P. Let the $x$-coordinate of point P be $k$.
What is the value of $\lim _ { n \rightarrow \infty } \frac { \left( \frac { a } { 3 } \right) ^ { n + k } } { \left( \frac { a } { 3 } \right) ^ { n + 1 } + 1 }$? [3 points]
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5

We define the function $\theta : \mathbb { R } \rightarrow \mathbb { C }$ by $$\begin{cases} \theta ( x ) = 0 & \text { if } x \leqslant 0 \\ \theta ( x ) = \exp \left( - \frac { \ln ^ { 2 } x } { 4 \pi ^ { 2 } } + \mathrm { i } \frac { \ln x } { 2 \pi } \right) & \text { if } x > 0 \end{cases}$$
Show that $\lim _ { \substack { x \rightarrow 0 \\ x > 0 } } | \theta ( x ) | = 0$.

$\lim _ { x \rightarrow 2 } \frac { 3 ^ { x } + 3 ^ { 3 - x } - 12 } { 3 ^ { - \frac { x } { 2 } } - 3 ^ { 1 - x } }$ is equal to

If $\alpha = \lim _ { x \rightarrow 0 ^ { + } } \left( \frac { \mathrm { e } ^ { \sqrt { \tan x } } - \mathrm { e } ^ { \sqrt { x } } } { \sqrt { \tan x } - \sqrt { x } } \right)$ and $\beta = \lim _ { x \rightarrow 0 } ( 1 + \sin x ) ^ { \frac { 1 } { 2 } \cot x }$ are the roots of the quadratic equation $a x ^ { 2 } + b x - \sqrt { \mathrm { e } } = 0$, then $12 \log _ { \mathrm { e } } ( \mathrm { a } + \mathrm { b } )$ is equal to $\_\_\_\_$

$\lim _ { x \rightarrow 0 } \frac { e - ( 1 + 2 x ) ^ { \frac { 1 } { 2 x } } } { x }$ is equal to
(1) 0
(2) $\frac { - 2 } { e }$
(3) e
(4) $e - e ^ { 2 }$

If $\lim _ { x \rightarrow \infty } \left( \left( \frac { \mathrm { e } } { 1 - \mathrm { e } } \right) \left( \frac { 1 } { \mathrm { e } } - \frac { x } { 1 + x } \right) \right) ^ { x } = \alpha$, then the value of $\frac { \log _ { \mathrm { e } } \alpha } { 1 + \log _ { \mathrm { e } } \alpha }$ equals:
(1) $e ^ { - 1 }$
(2) $\mathrm { e } ^ { 2 }$
(3) $e ^ { - 2 }$
(4) e