Find the value of $\lim _ { x \rightarrow 1 } \frac { ( x - 1 ) \left( x ^ { 2 } + 3 x + 7 \right) } { x - 1 }$. [3 points]

For an application $f : \mathbb{R}_+^* \rightarrow \mathbb{R}$ of class $\mathcal{C}^\infty$, we define the application $$\delta(f) : \left\{ \begin{array}{l} \mathbb{R}_+^* \rightarrow \mathbb{R} \\ x \mapsto f(x+1) - f(x) \end{array} \right.$$
Explain why, for every $x > 0$, there exists a $y_1 \in \left]0, 1\right[$ such that $$(\delta(f))(x) = f'(x + y_1)$$

The limit $\lim \left[ \left\{ 1 - \cos \left( \sin ^ { 2 } a x \right) \right\} / x \right]$ as $x -> 0$
(a) Equals 1
(b) Equals a
(c) Equals 0
(d) Does not exist

Let $m$ and $n$ be two positive integers greater than 1 . If
$$\lim _ { \alpha \rightarrow 0 } \left( \frac { e ^ { \cos \left( \alpha ^ { n } \right) } - e } { \alpha ^ { m } } \right) = - \left( \frac { e } { 2 } \right)$$
then the value of $\frac { m } { n }$ is

Let $\alpha, \beta \in \mathbb{R}$ be such that $\lim_{x \rightarrow 0} \frac{x^2 \sin(\beta x)}{\alpha x - \sin x} = 1$. Then $6(\alpha + \beta)$ equals

Let $k \in \mathbb { R }$. If $\lim _ { x \rightarrow 0 + } ( \sin ( \sin k x ) + \cos x + x ) ^ { \frac { 2 } { x } } = e ^ { 6 }$, then the value of $k$ is
(A) 1
(B) 2
(C) 3
(D) 4

$\lim _ { x \rightarrow 0 } \frac { \sin \left( \pi \cos ^ { 2 } x \right) } { x ^ { 2 } }$ is equal to
(1) $- \pi$
(2) $\pi$
(3) $\frac { \pi } { 2 }$
(4) 1

$\lim _ { x \rightarrow 0 } \frac { ( 1 - \cos 2 x ) ^ { 2 } } { 2 x \tan x - x \tan 2 x }$ is
(1) 2
(2) $- \frac { 1 } { 2 }$
(3) $- 2$
(4) $\frac { 1 } { 2 }$

$\lim_{x \to \frac{\pi}{2}} \dfrac{\cot x - \cos x}{(\pi - 2x)^3}$ equals
(1) $\dfrac{1}{24}$
(2) $\dfrac{1}{16}$
(3) $\dfrac{1}{8}$
(4) $\dfrac{1}{4}$

$\lim _ { x \rightarrow 0 } \frac { ( 27 + x ) ^ { \frac { 1 } { 3 } } - 3 } { 9 - ( 27 + x ) ^ { \frac { 2 } { 3 } } }$ equals
(1) $- \frac { 1 } { 6 }$
(2) $\frac { 1 } { 6 }$
(3) $\frac { 1 } { 3 }$
(4) $- \frac { 1 } { 3 }$

$\lim _ { x \rightarrow 1 ^ { - } } \frac { \sqrt { \pi } - \sqrt { 2 \sin ^ { - 1 } x } } { \sqrt { 1 - x } }$ is equal to
(1) $\sqrt { \pi }$
(2) $\sqrt { \frac { 2 } { \pi } }$
(3) $\frac { 1 } { \sqrt { 2 \pi } }$
(4) $\sqrt { \frac { \pi } { 2 } }$

If $\lim _ { x \rightarrow 0 } \frac { a x - \left( e ^ { 4 x } - 1 \right) } { a x \left( e ^ { 4 x } - 1 \right) }$ exists and is equal to $b$, then the value of $a - 2 b$ is $\underline{\hspace{1cm}}$.

If $\lim _ { x \rightarrow 0 } \frac { a e ^ { x } - b \cos x + c e ^ { - x } } { x \sin x } = 2$, then $a + b + c$ is equal to $\_\_\_\_$.

If $\lim _ { x \rightarrow 0 } \left[ \frac { \alpha x e ^ { x } - \beta \log _ { e } ( 1 + x ) + \gamma x ^ { 2 } e ^ { - x } } { x \sin ^ { 2 } x } \right] = 10 , \alpha , \beta , \gamma \in R$, then the value of $\alpha + \beta + \gamma$ is $\underline{\hspace{1cm}}$.

$\lim _ { x \rightarrow 0 } \frac { \sin ^ { 2 } \left( \pi \cos ^ { 4 } x \right) } { x ^ { 4 } }$ is equal to :
(1) $2 \pi ^ { 2 }$
(2) $\pi ^ { 2 }$
(3) $4 \pi ^ { 2 }$
(4) $4 \pi$

$\lim _ { x \rightarrow 0 } \frac { \cos ( \sin x ) - \cos x } { x ^ { 4 } }$ is equal to
(1) $\frac { 1 } { 3 }$
(2) $\frac { 1 } { 6 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 1 } { 12 }$

$\lim_{x \to \frac{\pi}{2}} \tan^2 x \left[(2\sin^2 x + 3\sin x + 4)^{\frac{1}{2}} - (\sin^2 x + 6\sin x + 2)^{\frac{1}{2}}\right]$ is equal to
(1) $\frac{1}{12}$
(2) $-\frac{1}{18}$
(3) $-\frac{1}{12}$
(4) $\frac{1}{6}$

$\lim_{x \rightarrow \frac{\pi}{4}} \frac{8\sqrt{2} - (\cos x + \sin x)^7}{\sqrt{2} - \sqrt{2}\sin 2x}$ is equal to
(1) 14
(2) 7
(3) $14\sqrt{2}$
(4) $7\sqrt{2}$

Let $x = 2$ be a root of the equation $x ^ { 2 } + p x + q = 0$ and $f ( x ) = \left\{ \begin{array} { c l } \frac { 1 - \cos \left( x ^ { 2 } - 4 p x + q ^ { 2 } + 8 q + 16 \right) } { ( x - 2 p ) ^ { 4 } } , & x \neq 2 p \\ 0 , & x = 2 p \end{array} \right.$. Then $\lim _ { x \rightarrow 2 p ^ { + } } [ f ( x ) ]$ where $[ \cdot ]$ denotes greatest integer function, is
(1) 2
(2) 1
(3) 0
(4) $- 1$

$\lim _ { x \rightarrow 0 } \left( \left( \frac { 1 - \cos ^ { 2 } ( 3 x ) } { \cos ^ { 3 } ( 4 x ) } \right) \left( \frac { \sin ^ { 3 } ( 4 x ) } { \left( \log _ { e } ( 2 x + 1 ) \right) ^ { 5 } } \right) \right)$ is equal to
(1) 15
(2) 9
(3) 18
(4) 24

$\lim _ { x \rightarrow 0 } \frac { e ^ { 2 \sin x } - 2 \sin x - 1 } { x ^ { 2 } }$
(1) is equal to $-1$
(2) does not exist
(3) is equal to 1
(4) is equal to 2

If $a = \lim _ { x \rightarrow 0 } \frac { \sqrt { 1 + \sqrt { 1 + x ^ { 4 } } } - \sqrt { 2 } } { x ^ { 4 } }$ and $b = \lim _ { x \rightarrow 0 } \frac { \sin ^ { 2 } x } { \sqrt { 2 } - \sqrt { 1 + \cos x } }$, then the value of $a b ^ { 3 }$ is:
(1) 36
(2) 32
(3) 25
(4) 30

If $\lim _ { x \rightarrow 0 } \frac { 3 + \alpha \sin x + \beta \cos x + \log _ { e } ( 1 - x ) } { 3 \tan ^ { 2 } x } = \frac { 1 } { 3 }$, then $2 \alpha - \beta$ is equal to :
(1) 2
(2) 7
(3) 5
(4) 1

Let $\mathrm { a } > 0$ be a root of the equation $2 x ^ { 2 } + x - 2 = 0$. If $\lim _ { x \rightarrow \frac { 1 } { \mathrm { a } } } \frac { 16 \left( 1 - \cos \left( 2 + x - 2 x ^ { 2 } \right) \right) } { ( 1 - \mathrm { a } x ) ^ { 2 } } = \alpha + \beta \sqrt { 17 }$, where $\alpha , \beta \in Z$, then $\alpha + \beta$ is equal to $\_\_\_\_$

The value of $\lim _ { x \rightarrow 0 } 2 \left( \frac { 1 - \cos x \sqrt { \cos 2 x } \sqrt [ 3 ] { \cos 3 x } \ldots \ldots \sqrt [ 10 ] { \cos 10 x } } { x ^ { 2 } } \right)$ is $\_\_\_\_$