kak tlong bantu saya untk jawab soal limit di atas

1.
[tex]lim \: \: x -> 4 \: \: \: \: \frac{4 - x}{2 - \sqrt{x} } = \frac{4 - x}{2 - \sqrt{x} } \times \frac{2 + \sqrt{x} }{2 + \sqrt{x} } \\ \frac{(4 - x)(2 + \sqrt{x)} }{4 - x} \\ lim \: x - > 4 \: \: \: \: \: 2 + \sqrt{x} = 2 + \sqrt{4} = 2 + 2 = 4[/tex]


2.
[tex]lim \: \: x -> 1 \: \: \: \frac{2 {x}^{2} - x - 1 }{3 {x}^{2} - x - 2 } = \frac{(2x + 1)(x - 1)}{(3x + 2)(x - 1)} = \frac{2x + 1}{3x + 2} \\ \frac{2(1) + 1}{3(1 )+ 2} = \frac{3}{5} [/tex]

3.
[tex]lim \: \: x - > 3 \: \: \: \: \frac{3 {x}^{2} - 5x - 12 }{x ^{2} - 9 } = \frac{(3x + 4)(x - 3)}{( x + 3)(x - 3)} \\ lim \: \: x - > 3 \: \: \: \: \frac{3x + 4}{x + 3} = \frac{3(3) + 4}{3 + 3} = \frac{12}{6} = 2[/tex]
4.
[tex]lim \: \: x - > 0 \: \: \: \: \frac{x + \sqrt{x} }{ \sqrt{x} } = \frac{x + \sqrt{x} }{ \sqrt{x} } \times \frac{ \sqrt{x} }{ \sqrt{x} } = \frac{x \sqrt{x} + x}{x} = \frac{x( \sqrt{x} + 1)}{x} \\ lim \: \: x - > 0 \: \: \: \: \sqrt{x} + 1 = 0 + 1 = 1[/tex]