Nilai lim_(x→2)⁡ (4-x^2)/(3-√(x^2+5))=⋯

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Bahas Soal Matematika   »   Limit   ›  

Nilai \( \displaystyle \lim_{x \to 2} \ \frac{4-x^2}{3-\sqrt{x^2+5}} = \cdots \)

  1. -1
  2. 0
  3. 2
  4. 6
  5. 8

(SPMB 2005)

Pembahasan:

\begin{aligned} \lim_{x \to 2} \ \frac{4-x^2}{3-\sqrt{x^2+5}} &= \lim_{x \to 2} \ \frac{4-x^2}{3-\sqrt{x^2+5}} \times \frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}} \\[8pt] &= \lim_{x \to 2} \ \frac{(4-x^2)(3+\sqrt{x^2+5})}{9-(x^2+5)} \\[8pt] &= \lim_{x \to 2} \ \frac{(4-x^2)(3+\sqrt{x^2+5})}{4-x^2} \\[8pt] &= \lim_{x \to 2} \ (3+\sqrt{x^2+5}) = 3+\sqrt{2^2+5} \\[8pt] &= 3 + 3 = 6 \end{aligned}

Jawaban D.