Question

mohon bantuannya, sertakan juga langkah penyelesaiannya​

Answer 1

2 [D.]

Penjelasan:

[tex]\sf \lim_{x \to \infty} \frac{2}{x} cot\frac{1}{x} \\\\=2\cdot \lim_{x \to \infty} \frac{1}{x} cot(\frac{1}{x})\\\\=2\cdot \lim_{x \to \infty} \frac{cot(\frac{1}{x})}{x}[/tex]  Gunakan identitas trigonometri:   [tex]\sf cot\:\alpha =\frac{cos\:\alpha }{sin\:\alpha }[/tex]

[tex]\sf =2\cdot \lim_{x \to \infty} \frac{cos(\frac{1}{x})}{x\cdot sin(\frac{1}{x})}\\\\=2\cdot \frac{\lim_{x \to \infty}cos(\frac{1}{x})}{\lim_{x \to \infty}x\cdot sin(\frac{1}{x})}[/tex]

[tex]\sf\lim_{x \to \infty}cos(\frac{1}{x}) =cos(\frac{1}{\infty} )=cos(0)=1[/tex]

[tex]\sf\lim_{x \to \infty}x\cdot sin(\frac{1}{x})=\infty\cdot sin(\frac{1}{\infty} )=\infty\cdot sin(0)=\infty\cdot 0[/tex]

'∞ · 0' Maka berlaku aturan L'Hopital (turunannya).

[tex]\sf\lim_{x \to \infty}x\cdot sin(\frac{1}{x})\\\\=\lim_{x \to \infty}cos (\frac{1}{x} )\\\\=cos(\frac{1}{\infty} )\\\\=cos(0)\\\\=1[/tex]

[tex]\sf \lim_{x \to \infty} \frac{2}{x} cot\frac{1}{x}\\\\=2\cdot \frac{\lim_{x \to \infty}cos(\frac{1}{x})}{\lim_{x \to \infty}x\cdot sin(\frac{1}{x})}\\\\=2\cdot \frac{1}{1} \\\\=2[/tex]

Jadi, hasilnya 2 [D.]

Answer 2

D. 2

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