What is the value of $\lim _ { x \rightarrow 0 } \frac { e ^ { x } - 1 } { 5 x }$? [2 points]
(1) 5
(2) $e$
(3) 1
(4) $\frac { 1 } { e }$
(5) $\frac { 1 } { 5 }$

As shown in the figure, in the coordinate plane, the circle $x ^ { 2 } + y ^ { 2 } = 1$ and the curve $y = \ln ( x + 1 )$ meet at point A in the first quadrant. For point $\mathrm { B } ( 1,0 )$, let H be the foot of the perpendicular from point P on arc AB to the $y$-axis, and let Q be the intersection of segment PH and the curve $y = \ln ( x + 1 )$. Let $\angle \mathrm { POB } = \theta$. If $S ( \theta )$ is the area of triangle OPQ and $L ( \theta )$ is the length of segment HQ, and $\lim _ { \theta \rightarrow + 0 } \frac { S ( \theta ) } { L ( \theta ) } = k$, find the value of $60 k$. (Here, $0 < \theta < \frac { \pi } { 6 }$ and O is the origin.) [4 points]

What is the value of $\lim _ { x \rightarrow 0 } \frac { e ^ { 6 x } - 1 } { \ln ( 1 + 3 x ) }$? [2 points]
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5

What is the value of $\lim _ { x \rightarrow 0 } \frac { x ^ { 2 } + 5 x } { \ln ( 1 + 3 x ) }$? [2 points]
(1) $\frac { 7 } { 3 }$
(2) 2
(3) $\frac { 5 } { 3 }$
(4) $\frac { 4 } { 3 }$
(5) 1

What is the value of $\lim _ { x \rightarrow 0 } \frac { 6 x } { e ^ { 4 x } - e ^ { 2 x } }$? [2 points]
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5

Find the value of $\lim_{x \rightarrow 0} \frac{\ln(1+3x)}{\ln(1+5x)}$. [2 points]
(1) $\frac{1}{5}$
(2) $\frac{2}{5}$
(3) $\frac{3}{5}$
(4) $\frac{4}{5}$
(5) 1

We consider a real $\alpha$ such that for every prime number $p$, $p^\alpha$ is a natural number. We apply relation $$\sum_{j=0}^{n} (-1)^{n-j} \binom{n}{j} f(x+j) = f^{(n)}(x + y_n) \quad \text{(IV.1)}$$ to the function $f_\alpha(x) = x^\alpha$ and to the integer $n = \lfloor \alpha \rfloor + 1$. The notations are those of question IV.A.4.
What is the limit of the expression $f_\alpha^{(n)}(x + y_n)$ when $x \in \mathbb{N}^*$ tends to $+\infty$?

We define the function $\varphi : \mathbb { R } \rightarrow \mathbb { R }$ by $$\begin{cases} \varphi ( x ) = \exp \left( \frac { - x } { \sqrt { 1 - x } } \right) & \text { if } x < 1 \\ \varphi ( x ) = 0 & \text { if } x \geqslant 1 \end{cases}$$
Deduce $\lim _ { \substack { x \rightarrow 1 \\ x < 1 } } \varphi ^ { ( p ) } ( x )$ for $p \in \mathbb { N } ^ { * }$.

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

$\lim_{x \to 2} \left[\frac{e^{x^{2}} - e^{2x}}{(x-2)e^{2x}}\right]$ equals
(a) 0
(b) 1
(c) 2
(d) 3

The limit $$\lim _ { x \rightarrow 0 } \frac { \left( e ^ { x } - 1 \right) \tan ^ { 2 } x } { x ^ { 3 } }$$ (A) does not exist
(B) exists and equals 0
(C) exists and equals $2/3$
(D) exists and equals 1

$\lim _ { x \rightarrow 0 } \frac { \left( e ^ { x } - 1 \right) \tan ^ { 2 } x } { x ^ { 3 } }$
(a) does not exist
(b) exists and equals 0
(c) exists and equals $\frac { 2 } { 3 }$
(d) exists and equals 1.

$\lim _ { x \rightarrow 0 } \frac { \left( e ^ { x } - 1 \right) \tan ^ { 2 } x } { x ^ { 3 } }$
(a) does not exist
(b) exists and equals 0
(c) exists and equals $\frac { 2 } { 3 }$
(d) exists and equals 1.

Let $\alpha$ be a positive real number. Let $f : \mathbb { R } \rightarrow \mathbb { R }$ and $g : ( \alpha , \infty ) \rightarrow \mathbb { R }$ be the functions defined by
$$f ( x ) = \sin \left( \frac { \pi x } { 12 } \right) \quad \text { and } \quad g ( x ) = \frac { 2 \log _ { \mathrm { e } } ( \sqrt { x } - \sqrt { \alpha } ) } { \log _ { \mathrm { e } } \left( e ^ { \sqrt { x } } - e ^ { \sqrt { \alpha } } \right) }$$
Then the value of $\lim _ { x \rightarrow \alpha ^ { + } } f ( g ( x ) )$ is $\_\_\_\_$.

Let $x _ { 0 }$ be the real number such that $e ^ { x _ { 0 } } + x _ { 0 } = 0$. For a given real number $\alpha$, define
$$g ( x ) = \frac { 3 x e ^ { x } + 3 x - \alpha e ^ { x } - \alpha x } { 3 \left( e ^ { x } + 1 \right) }$$
for all real numbers $x$.
Then which one of the following statements is TRUE?

(A)	For $\alpha = 2 , \lim _ { x \rightarrow x _ { 0 } } \left| \frac { g ( x ) + e ^ { x _ { 0 } } } { x - x _ { 0 } } \right| = 0$
(B)	For $\alpha = 2 , \lim _ { x \rightarrow x _ { 0 } } \left| \frac { g ( x ) + e ^ { x _ { 0 } } } { x - x _ { 0 } } \right| = 1$
(C)	For $\alpha = 3 , ~ \lim _ { x \rightarrow x _ { 0 } } \left| \frac { g ( x ) + e ^ { x _ { 0 } } } { x - x _ { 0 } } \right| = 0$
(D)	For $\alpha = 3 , ~ \lim _ { x \rightarrow x _ { 0 } } \left| \frac { g ( x ) + e ^ { x _ { 0 } } } { x - x _ { 0 } } \right| = \frac { 2 } { 3 }$

$$\lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{\ln x}$$
What is the value of this limit?
A) $\frac{-1}{2}$
B) $0$
C) $\frac{1}{2}$
D) $1$
E) $2$

$$\lim _ { x \rightarrow \infty } \frac { \ln ( x - 3 ) } { \ln \sqrt { x } }$$
What is the value of this limit?
A) 1
B) 2
C) 3
D) $\frac { 3 } { 2 }$
E) $\frac { 5 } { 2 }$