$$f ( x ) = \frac { x } { x + \sin x } \quad \text { and } \quad g ( x ) = \frac { x ^ { 4 } + x ^ { 6 } } { e ^ { x } - 1 - x ^ { 2 } }$$
(a) Limit as $x \rightarrow 0$ of $f ( x )$ is $\frac { 1 } { 2 }$.
(b) Limit as $x \rightarrow \infty$ of $f ( x )$ does not exist.
(c) Limit as $x \rightarrow \infty$ of $g ( x )$ is finite.
(d) Limit as $x \rightarrow 0$ of $g ( x )$ is 720.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Give a simple asymptotic equivalent of $f(x)$ as $x$ tends to $-1$.

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n }$ is defined as in Q19. We assume $k \in I _ { n } = \left\{ k \in \llbracket 0 , n \rrbracket \mid x _ { n , k } \in [ 0 , \ell + 1 ] \right\}$.
Deduce that $$B _ { n } \left( x _ { n , k } \right) = \frac { 1 } { \sqrt { 2 \pi } } \frac { 1 + O \left( \frac { 1 } { n } \right) } { \left( 1 - \frac { x _ { n , k } ^ { 2 } } { n } \right) ^ { \frac { n + 1 } { 2 } } \left( 1 + \frac { x _ { n , k } } { \sqrt { n } } \right) ^ { \frac { x _ { n , k } } { 2 } \sqrt { n } } \left( 1 - \frac { x _ { n , k } } { \sqrt { n } } \right) ^ { - \frac { x _ { n , k } } { 2 } \sqrt { n } } }$$ then that $$B _ { n } \left( x _ { n , k } \right) = \frac { 1 } { \sqrt { 2 \pi } } \exp \left( - \frac { x _ { n , k } ^ { 2 } } { 2 } \right) \left( 1 + O \left( \frac { 1 } { \sqrt { n } } \right) \right)$$

Let $\varphi$ be the function defined by
$$\forall t \in ] - 1,1 \left[ \backslash \{ 0 \} , \quad \varphi ( t ) = ( 1 - t ) ^ { 1 - 1 / t } \right.$$
Justify that $\varphi$ is extendable by continuity at 0 and specify the value of its extension at 0. We will still denote this extension by $\varphi$.

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

If $a _ { n } = \left( 1 + \frac { 1 } { n ^ { 2 } } \right) \left( 1 + \frac { 2 ^ { 2 } } { n ^ { 2 } } \right) ^ { 2 } \left( 1 + \frac { 3 ^ { 2 } } { n ^ { 2 } } \right) ^ { 3 } \cdots \left( 1 + \frac { n ^ { 2 } } { n ^ { 2 } } \right) ^ { n }$, then $$\lim _ { n \rightarrow \infty } a _ { n } ^ { - 1 / n ^ { 2 } }$$ is
(A) 0
(B) 1
(C) $e$
(D) $\sqrt { e } / 2$

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

If $a _ { n } = \left( 1 + \frac { 1 } { n ^ { 2 } } \right) \left( 1 + \frac { 2 ^ { 2 } } { n ^ { 2 } } \right) ^ { 2 } \left( 1 + \frac { 3 ^ { 2 } } { n ^ { 2 } } \right) ^ { 3 } \cdots \left( 1 + \frac { n ^ { 2 } } { n ^ { 2 } } \right) ^ { n }$, then $$\lim _ { n \rightarrow \infty } a _ { n } ^ { - 1 / n ^ { 2 } }$$ is
(A) 0
(B) 1
(C) $e$
(D) $\sqrt { e } / 2$

Let $f ( x ) = \frac { 1 } { 2 } x \sin x - ( 1 - \cos x )$. The smallest positive integer $k$ such that $\lim _ { x \rightarrow 0 } \frac { f ( x ) } { x ^ { k } } \neq 0$ is:
(A) 3
(B) 4
(C) 5
(D) 6.

For all natural numbers $n$, let $$A_{n} = \sqrt{2 - \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}} \quad (n \text{ many radicals})$$
(a) Show that for $n \geq 2$, $$A_{n} = 2\sin\frac{\pi}{2^{n+1}}$$
(b) Hence, or otherwise, evaluate the limit $$\lim_{n \rightarrow \infty} 2^{n} A_{n}$$

Let $$P(x) = 1 + 2x + 7x^2 + 13x^3, \quad x \in \mathbb{R}$$ Calculate for all $x \in \mathbb{R}$, $$\lim_{n \to \infty} \left(P\left(\frac{x}{n}\right)\right)^n$$

The limit $$\lim _ { n \rightarrow \infty } n ^ { - \frac { 3 } { 2 } } \left( ( n + 1 ) ^ { ( n + 1 ) } ( n + 2 ) ^ { ( n + 2 ) } \ldots ( 2 n ) ^ { ( 2 n ) } \right) ^ { \frac { 1 } { n ^ { 2 } } }$$ equals
(A) 0.
(B) 1.
(C) $e ^ { - \frac { 1 } { 4 } }$.
(D) $4 e ^ { - \frac { 3 } { 4 } }$.

Let
$$L = \lim _ { x \rightarrow 0 } \frac { a - \sqrt { a ^ { 2 } - x ^ { 2 } } - \frac { x ^ { 2 } } { 4 } } { x ^ { 4 } } , \quad a > 0 .$$
If $L$ is finite, then
(A) $\quad a = 2$
(B) $\quad a = 1$
(C) $\quad L = \frac { 1 } { 64 }$
(D) $\quad L = \frac { 1 } { 32 }$

If
$$\beta = \lim _ { x \rightarrow 0 } \frac { e ^ { x ^ { 3 } } - \left( 1 - x ^ { 3 } \right) ^ { \frac { 1 } { 3 } } + \left( \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } - 1 \right) \sin x } { x \sin ^ { 2 } x }$$
then the value of $6 \beta$ is $\_\_\_\_$ .

$\lim_{x \rightarrow \infty} \frac{(\sqrt{3x+1} + \sqrt{3x-1})^6 + (\sqrt{3x+1} - \sqrt{3x-1})^6}{\left(x + \sqrt{x^2-1}\right)^6 + \left(x - \sqrt{x^2-1}\right)^6} x^3$
(1) is equal to $\frac{27}{2}$
(2) is equal to 9
(3) does not exist
(4) is equal to 27

If $\lim _ { x \rightarrow 0 } \frac { e ^ { ax } - \cos ( bx ) - \frac { cx e ^ { cx } } { 2 } } { 1 - \cos ( 2 x ) } = 17$, then $5 a ^ { 2 } + b ^ { 2 }$ is equal to
(1) 64
(2) 72
(3) 68
(4) 76