Question

~QUIZ~
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Note: Soal terlampir.
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Syarat untuk menjawab soal :
● Dilarang jawaban berupa komentar spam atau asal²an.
● Dilarang copas jawaban dari google.
● Jawabannya harus disertai dengan penjelasan yang masuk akal.
● Gunakanlah kata-kata jawabanmu sendiri yang baik dan benar.

Answer 1

Verified answer

Jawaban:

[tex]B.-2[/tex]



Pengerjaan Soal:

[tex] terlampir [/tex]

Answer 2

Jawab:

-2 (B)

Penjelasan:

[tex]\lim\limits_{x \to 0} \frac{2sin(x)+sin(2x)}{-2x} \\[/tex]

Jika menggunakan cara substitusi,

[tex]\lim\limits_{x \to 0} \frac{2sin(x)+sin(2x)}{-2x} \\\\= \frac{2sin(0)+sin(2(0))}{-2(0)} \\\\=\frac{2(0)+0}{0} \\\\=\frac{0}{0}[/tex]

Hasilnya 0/0, maka berlaku aturan L'Hôpital, di mana:

[tex]\boxed{ \lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}}[/tex]

[tex]\lim\limits_{x \to 0} \frac{2sin(x)+sin(2x)}{-2x} \\\\=-\frac{1}{2} \cdot \lim\limits_{x \to 0} \frac{2sin(x)+sin(2x)}{x}\\\\=-\frac{1}{2} \cdot (\lim\limits_{x \to 0} \frac{2sin(x)}{x}+\lim\limits_{x \to 0} \frac{sin(2x)}{x})\\\\=-\frac{1}{2} \cdot (2\cdot \lim\limits_{x \to 0} \frac{sin(x)}{x} +\lim\limits_{x \to 0} \frac{sin(2x)}{x})[/tex]

Aplikasikan aturan L'Hôpital.

[tex]=-\frac{1}{2} \cdot (2\cdot \lim\limits_{x \to 0} \frac{cos(x)}{1} +\lim\limits_{x \to 0} \frac{2\cdot cos(2x)}{1})\\\\=-\frac{1}{2} \cdot (2\cdot \frac{cos(0)}{1} +\frac{2\cdot cos(2\cdot0)}{1} )\\\\=-\frac{1}{2} \cdot (2\cdot cos(0)+2\cdot cos(0) )\\\\=-\frac{1}{2} \cdot (2\cdot 1 +2\cdot 1 )\\\\=-\frac{2+2}{2} \\\\=-\frac{4}{2} \\\\\boxed{=-2}[/tex]