13 & 14 Tolong dijawab dan diberi penjelasan nilaix dapat dari mana aja supaya bisa dipahami.... Question from @Irwan7771

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Kelas: 11
Mapel: Matematika
Kategori: Limit
Kata kunci: Limit
Kode: 11.2.7 (Kelas 11 Matematika Bab 7-Limit)

13.
$\lim_{x \to \infty} 2x- \sqrt{4x^2+x+3} \\= \lim_{x \to \infty} 2x- \sqrt{4x^2+x+3} \times \frac{2x+ \sqrt{4x^2+x+3} }{2x+ \sqrt{4x^2+x+3} } \\ = \lim_{x \to \infty} \frac{4x^2-(4x^2+x+3)}{2x+ \sqrt{4x^2+x+3} } \\ = \lim_{x \to \infty} \frac{-x-3}{2x+ \sqrt{4x^2+x+3} } \times \frac{ \frac{1}{x} }{ \frac{1}{x} } \\ = \lim_{x \to \infty} \frac{-1- \frac{3}{x} }{2+ \sqrt{4+ \frac{1}{x}+ \frac{3}{x^2} } }$
$= \lim_{x \to \infty} \frac{-1- \frac{3}{\infty} }{2+ \sqrt{4+ \frac{1}{\infty}+ \frac{3}{\infty} } } \\ = \frac{-1-0}{2+ \sqrt{4+0+0} } \\ = \frac{-1}{2+2} \\ =- \frac{1}{4}$

14.
$\lim_{x \to 4} \frac{3- \sqrt{3x-3} }{x^2+x-20}\times \frac{3+ \sqrt{3x-3} }{3+ \sqrt{3x-3} } \\ = \lim_{x \to 4} \frac{9-(3x-3)}{(x-4)(x+5)(3+ \sqrt{3x-3}) } \\=lim_{x \to 4} \frac{12-3x}{(x-4)(x+5)(3+ \sqrt{3x-3}) } \\ =lim_{x \to 4} \frac{-3(x-4)}{(x-4)(x+5)(3+ \sqrt{3x-3}) } \\ =lim_{x \to 4} \frac{-3}{(x+5)(3+ \sqrt{3x-3}) } \\ = \frac{-3}{(4+5)(3+ \sqrt{3(4)-3}) } \\ = \frac{-3}{9(3+ \sqrt{9} )} \\ = \frac{-3}{9(3+3)} \\ = \frac{-3}{9(6)} \\ =- \frac{1}{18}$

Semangat belajar!
Semoga membantu :)

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Irwan7771 Makasih kak

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