Nilai lim_(x→3)⁡ (x^2-x-6)/(√(3x^2-2)-5)=⋯

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Bahas Soal Matematika   »   Limit   ›  

Nilai \( \displaystyle \lim_{x \to 3} \ \frac{x^2-x-6}{\sqrt{3x^2-2}-5} = \cdots \)

  1. 0
  2. 25/9
  3. 25/6
  4. 25/3

(UN SMA IPA 2019)

Pembahasan:

\begin{aligned} \lim_{x \to 3} \ \frac{x^2-x-6}{\sqrt{3x^2-2}-5} &= \lim_{x \to 3} \ \frac{x^2-x-6}{\sqrt{3x^2-2}-5} \times \frac{\sqrt{3x^2-2}+5}{\sqrt{3x^2-2}+5} \\[8pt] &= \lim_{x \to 3} \ \frac{(x^2-x-6)(\sqrt{3x^2-2}+5)}{(3x^2-2)-25} \\[8pt] &= \lim_{x \to 3} \ \frac{(x-3)(x+2)(\sqrt{3x^2-2}+5)}{3(x^2-9)} \\[8pt] &= \lim_{x \to 3} \ \frac{(x-3)(x+2)(\sqrt{3x^2-2}+5)}{3(x-3)(x+3)} \\[8pt] &= \lim_{x \to 3} \ \frac{(x+2)(\sqrt{3x^2-2}+5)}{3(x+3)} \\[8pt] &= \frac{(3+2)(\sqrt{3(3)^2-2}+5)}{3(3+3)} \\[8pt] &= \frac{(5)(\sqrt{25}+5)}{18} = \frac{(5)(10)}{18} = \frac{25}{9} \end{aligned}

Jawaban B.