Nilai \( \displaystyle \lim_{x \to 1} \ \frac{4t^4 + 4t - 72}{(t-2)(t^2+3t+2)} = \cdots \)
- \( \frac{11}{4} \)
- \( \frac{11}{3} \)
- \( 11 \)
- \( 22 \)
- \( 33 \)
(UMB PTN 2008)
Pembahasan:
\begin{aligned} \lim_{x \to 2} \ \frac{4t^4 + 4t - 72}{(t-2)(t^2+3t+2)} &= \lim_{x \to 2} \ \frac{4(t-2)(t^3+2t^2+4t+9)}{(t-2)(t^2+3t+2)} \\[8pt] &= \lim_{x \to 2} \ \frac{4(t^3+2t^2+4t+9)}{(t^2+3t+2)} \\[8pt] &= \frac{4((2)^3+(2)(2)^2+(4)(2)+9)}{(2)^2+(3)(2)+2} \\[8pt] &= \frac{4(33)}{12} = 11 \end{aligned}
Jawaban C.