Question

tolong bantu jawab beserta caranya kak ☺️ ​

Answer 1

▪︎[tex]\lim\limits_{x\to6} \frac{12-2x}{3-\sqrt{x+3}} =[/tex] 12

Opsi A

▪︎[tex]\lim\limits_{x\to3} \frac{x-3}{\sqrt{x^2-2}-\sqrt{7}} =[/tex]

[tex]\frac{\sqrt{7}}{3} [/tex]

Opsi E

[tex]\\\green{Diketahui:} [/tex]

[tex]\purple{Ditanya:}[/tex]

▪︎[tex]\lim\limits_{x\to6} \frac{12-2x}{3-\sqrt{x+3}} =[/tex] ?

▪︎[tex]\lim\limits_{x\to3} \frac{x-3}{\sqrt{x^2-2}-\sqrt{7}} =[/tex]?

[tex]\\\\\blue{Pembahasan:}\\[/tex]

[tex]\frac{12-2x}{3-\sqrt{x+3}} = [/tex]

[tex]\frac{12-2x}{3-\sqrt{x+3}}.\frac{3+\sqrt{x+3}}{3+\sqrt{x+3}}= [/tex]

[tex]\frac{(12-2x)(3+\sqrt{x+3})}{9-(x+3)}=[/tex]

[tex]2\frac{(6-x)(3+\sqrt{x+3})}{(6-x)}=[/tex]

[tex]2(3+\sqrt{x+3})[/tex]

[tex]\lim\limits_{x\to6} \frac{12-2x}{3-\sqrt{x+3}} =[/tex]

[tex]\lim\limits_{x\to6} 2(3+\sqrt{x+3})=[/tex]

[tex] 2(3+\sqrt{6+3})=[/tex]

[tex] 2(3+\sqrt{9})=[/tex]

2(3 + 3) =

2(6) =

12

[tex]\\[/tex]

[tex]\frac{x-3}{\sqrt{x^2-2}-\sqrt{7}} =[/tex]

[tex]\frac{(x-3)(\sqrt{x^2-2}+\sqrt{7})}{(\sqrt{x^2-2}-\sqrt{7})(\sqrt{x^2-2}+\sqrt{7})} =[/tex]

[tex]\frac{(x-3)(\sqrt{x^2-2}+\sqrt{7})}{x^2-2-7} =[/tex]

[tex]\frac{(x-3)(\sqrt{x^2-2}+\sqrt{7})}{x^2-9} =[/tex]

[tex]\frac{(x-3)(\sqrt{x^2-2}+\sqrt{7})}{(x+3)(x-3)} =[/tex]

[tex]\frac{(\sqrt{x^2-2}+\sqrt{7})}{(x+3)} =[/tex]

[tex]\lim\limits_{x\to3} \frac{x-3}{\sqrt{x^2-2}-\sqrt{7}} =[/tex]

[tex]\lim\limits_{x\to3} \frac{(\sqrt{x^2-2}+\sqrt{7})}{(x+3)} =[/tex]

[tex]\frac{(\sqrt{3^2-2}+\sqrt{7})}{(3+3)} =[/tex]

[tex]\frac{\sqrt{7} + \sqrt{7}}{6} = [/tex]

[tex]\frac{2\sqrt{7}}{6} = [/tex]

[tex]\frac{\sqrt{7}}{3} [/tex]

[tex]\\\\\blue{Pelajari~lebih~ lanjut:}[/tex]

MATERI Soal Limit:

•https://brainly.co.id/tugas/257129

•https://brainly.co.id/tugas/14435632

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[tex]\\\\\blue{Detail~ Jawaban:}[/tex]

[tex]\bullet[/tex]Mapel : Matematika

[tex]\bullet[/tex] Kelas : 11

[tex]\bullet[/tex] Materi : Bab 7 Limit

[tex]\bullet[/tex] Kata Kunci : limit

[tex]\bullet[/tex]Kode soal : 2

[tex]\bullet[/tex] Kode kategori : 11.2.7