Soal tentang limit..........

Soal tentang limit……….

Jawab:

-5/6

Penjelasan dengan langkah-langkah:

$lim_{x to -3} frac{5}{x+3}+frac{30}{x^2-9}\\=lim_{x to -3} frac{5}{x+3}+frac{30}{(x+3)(x-3)}\\=lim_{x to -3} frac{5(x-3)}{(x+3)(x-3)}+frac{30}{(x+3)(x-3)}\\=lim_{x to -3} frac{5x-15+30}{(x+3)(x-3)}\\=lim_{x to -3} frac{5x+15}{(x+3)(x-3)}\\=lim_{x to -3} frac{5(x+3)}{(x+3)(x-3)}\\=lim_{x to -3} frac{5}{x-3}\\=frac{5}{-3-3}\\=-frac{5}{6}$

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Mapel: Matematika

Kelas : 11

Bab : Limit Fungsi Aljabar

Kata Kunci : limit, aljabar, substitusi

Kode Kategorisasi: 11.2.8