Nilai lim_(x→3)⁡ (x-3)/(3-√(x+6))=⋯

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Bahas Soal Matematika   »   Limit   ›  

Nilai \( \displaystyle \lim_{x \to 3} \ \frac{x-3}{3-\sqrt{x+6}} = \cdots \)

  1. -6
  2. -3
  3. 0
  4. 3
  5. 6

(UMB PTN 2012)

Pembahasan:

\begin{aligned} \lim_{x \to 3} \ \frac{x-3}{3-\sqrt{x+6}} &= \lim_{x \to 3} \ \frac{x-3}{3-\sqrt{x+6}} \times \frac{3+\sqrt{x+6}}{3+\sqrt{x+6}} \\[8pt] &= \lim_{x \to 3} \ \frac{(x-3)(3+\sqrt{x+6})}{9-(x+6)} \\[8pt] &= \lim_{x \to 3} \ \frac{(x-3)(3+\sqrt{x+6})}{9-(x+6)} \\[8pt] &= \lim_{x \to 3} \ \frac{(x-3)(3+\sqrt{x+6})}{-(x-3)} \\[8pt] &= \lim_{x \to 3} \ \frac{(3+\sqrt{x+6})}{-1} \\[8pt] &= \frac{3+\sqrt{3+6}}{-1} = -6 \end{aligned}

Jawaban A.