Nilai lim_(x→4)⁡ (x^2-16)/(5-√(x^2+9))=⋯

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Bahas Soal Matematika   »   Limit   ›  

Nilai \( \displaystyle \lim_{x \to 4} \ \frac{x^2-16}{5-\sqrt{x^2+9}} = \cdots \)

  1. -20
  2. -10
  3. 0
  4. 8
  5. 20

(SPMB 2007)

Pembahasan:

\begin{aligned} \lim_{x \to 4} \ \frac{x^2-16}{5-\sqrt{x^2+9}} &= \lim_{x \to 4} \ \frac{x^2-16}{5-\sqrt{x^2+9}} \times \frac{5+\sqrt{x^2+9}}{5+\sqrt{x^2+9}} \\[8pt] &= \lim_{x \to 4} \ \frac{(x^2-16)(5+\sqrt{x^2+9})}{25-(x^2+9)} \\[8pt] &= \lim_{x \to 4} \ \frac{(x^2-16)(5+\sqrt{x^2+9})}{-(x^2-16)} \\[8pt] &= \lim_{x \to 4} \ -(5+\sqrt{x^2+9}) \\[8pt] &= -(5+\sqrt{4^2+9}) \\[8pt] &= -10 \end{aligned}

Jawaban B.