Nilai lim_(x→2)⁡ (x^3-8)/(x^2+x-6)=⋯

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Bahas Soal Matematika   »   Limit   ›  

Nilai \( \displaystyle \lim_{x \to 2} \ \frac{x^3-8}{x^2+x-6} = \cdots \)

  1. \( \frac{3}{4} \)
  2. \( \frac{2}{15} \)
  3. \( 1 \frac{1}{3} \)
  4. \( 2 \frac{2}{5} \)
  5. \( 6 \)

(SPMB 2004)

Pembahasan:

\begin{aligned} \lim_{x \to 2} \ \frac{x^3-8}{x^2+x-6} &= \lim_{x \to 2} \ \frac{(x-2)(x^2+2x+4)}{(x-2)(x+3)} \\[8pt] &= \lim_{x \to 2} \ \frac{x^2+2x+4}{x+3} \\[8pt] &= \frac{2^2 + 2(2) + 4}{2 + 3} \\[8pt] &= \frac{12}{5} = 2 \frac{2}{5} \end{aligned}

Jawaban D.