Nilai dari lim x mendekati 1 tan (x-1) sin (1-√x) dibagi x²-2x+1 atau [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] adalah [tex]-\frac{1}{2}[/tex]
Untuk menjawab soal di atas, kamu dapat menggunakan rumus limit trigonometri.
Rumus limit trigonometri:
- [tex]\lim_{x \to 0} \frac{\sin ax}{bx}[/tex] = [tex]\lim_{x \to 0} \frac{\tan ax}{bx}[/tex] = [tex]\frac{a}{b}[/tex]
- [tex]\lim_{x \to 0} \frac{ax}{\sin bx}[/tex] = [tex]\lim_{x \to 0} \frac{ax}{\tan bx}[/tex] = [tex]\frac{a}{b}[/tex]
- [tex]\lim_{x \to a} \frac{\sin (x-a)}{(x-a)}[/tex] = [tex]\lim_{x \to a} \frac{\tan (x-a)}{(x-a)}[/tex] = 1
Penjelasan dengan langkah-langkah:
Diketahui:
[tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex]
Ditanya:
Nilai dari [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] adalah
Jawab:
Gunakan rumus yang ke tiga:
⇔ [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] = [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{(x-1)(x-1)}[/tex]
⇒ [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] = [tex]\lim_{x \to 1} \frac{\tan (x-1)}{(x-1)} \times \lim_{x \to 1} \frac{\sin (1-\sqrt{x})} {(x-1)}[/tex]
⇒ [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] = [tex]1 \times \lim_{x \to 1} \frac{\sin (1-\sqrt{x})} {(\sqrt{x} -1)(\sqrt{x} +1)}[/tex]
⇒ [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] = [tex]1 \times \lim_{x \to 1} \frac{\sin (1-\sqrt{x})} {-(1-\sqrt{x} )(\sqrt{x} +1)}[/tex]
⇒ [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] = [tex]1 \times \lim_{x \to 1} \frac{\sin (1-\sqrt{x})} {(1-\sqrt{x} )} \times \frac{-1}{(\sqrt{x} +1)}[/tex]
⇒ [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] = [tex]1 \times 1 \times \lim_{x \to 1} \frac{-1}{(\sqrt{x} +1)}[/tex]
⇒ [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] = [tex]1 \times 1 \times \frac{-1}{(\sqrt{1} +1)}[/tex]
⇒ [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] = [tex]1 \times 1 \times \frac{-1}{(2)}[/tex]
⇒ [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] = [tex]-\frac{1}{2}[/tex]
Jadi, nilai dari [tex]\lim_{x \to 1} \frac{\tan (x-1) \sin (1-\sqrt{x} )}{x^2 - 2x+1}[/tex] adalah [tex]-\frac{1}{2}[/tex]
Pelajari lebih lanjut:
- Bab limit, .... berapa nilai limit berikut, ....... : https://brainly.co.id/tugas/30489523
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