Найти приделы функции

1.а.1 при х ->к

$\frac{4 - x}{ {x}^{2} - {4}^{2} } = \frac{4 - x}{(x - 4)(x+ 4)} = \frac{ - 1}{x + 4} = - \frac{1}{8}$

при х -> беск

1 а 2

$\frac{4 - x}{ {x}^{2} - 16} = \frac{ \frac{4}{ {x}^{2} - } - \frac{x}{ {x}^{2} } }{ \frac{ {x}^{2} }{ {x}^{2} } - \frac{16}{ {x}^{2} } } = \frac{0 - 0}{1 - 0} = 0$

1б

$\frac{ \sin(5x) }{arc \sin(4x) } = \\ \ \frac{ \sin(5x) \times 5x}{5x} \times \frac{1}{arcsin(4x)} = \\ \frac{1 \times 5x}{1} \times \frac{4}{4arcsin4x} = \frac{5}{4}$

$( \frac{ {x}^{2} }{ {x}^{2} - 4x} ) ^{5x} = \frac{ {x}^{2} - 4x + 4x }{ {x}^{2} - 4x } = \\ (1 + 4 \times \frac{1}{x - 4} ) ^{5x \times \frac{x - 4}{ x- 4} } = e ^{4 \times lim \frac{5x}{x - 4} } = e ^{4}$