Matematika soal:1. limit = 3x/2-√4+x x=> 02. limit = 8x /4-√16+x x => 03.limit = 4x / 5- √ 25 + x x=> 04. limit = x-2 / 2-√2x x=> 25. limit = x-3/ 3-√3x x=> 36. limit = √x-3 / x-9 x=> 9Jalannya bisa di liat di gambarTolong bantu jawab ya teman-teman​

melvinagislasonulv – March 09, 2023

soal:

1. limit = 3x/2-√4+x
x=> 0

2. limit = 8x /4-√16+x
x => 0

3.limit = 4x / 5- √ 25 + x
x=> 0

4. limit = x-2 / 2-√2x
x=> 2

5. limit = x-3/ 3-√3x
x=> 3

6. limit = √x-3 / x-9
x=> 9

Jalannya bisa di liat di gambar

Tolong bantu jawab ya teman-teman​

Jawaban:

1.

[tex]limit_{x = 0} = \frac{3x}{2 - \sqrt{4 + x} } \\ limit_{x = 0} = \frac{3x}{2 - \sqrt{4 + x} } \times \frac{2 + \sqrt{4 + x} }{2 + \sqrt{4 + x} } \\ limit_{x = 0} = \frac{6x + 3x \sqrt{4 + x} }{ - x} \\ limit_{x = 0} = - 6 - 3 \sqrt{4 + x} \\ = - 6 - 3 \sqrt{4} \\ = - 12[/tex]

2.

[tex]limit_{x = 0} = \frac{8x}{4 - \sqrt{16 + x} } \\ limit_{x = 0} = \frac{8x}{4 - \sqrt{16 + x} } \times \frac{4 + \sqrt{16 + x} }{4 + \sqrt{16 + x} } \\ limit_{x = 0} = \frac{32x + 8x \sqrt{16 + x} }{ - x} \\ limit_{x = 0} = - 32 - 8 \sqrt{16 + x} \\ = - 32 - 8 \sqrt{16 } \\ = - 64[/tex]

begitu soal seterusnya

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