Nilai \( \displaystyle \lim_{x \to 0} \ \frac{5x^2+x}{\sqrt{4+x}-2} = \cdots \)
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(SPMB 2005)
Pembahasan:
\begin{aligned} \lim_{x \to 0} \ \frac{5x^2+x}{\sqrt{4+x}-2} &= \lim_{x \to 0} \ \frac{5x^2+x}{\sqrt{4+x}-2} \times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2} \\[8pt] &= \lim_{x \to 0} \ \frac{(5x^2+x)(\sqrt{4+x}+2)}{(4+x)-4} \\[8pt] &= \lim_{x \to 0} \ \frac{x(5x+1)(\sqrt{4+x}-2)}{x} \\[8pt] &= \lim_{x \to 0} \ (5x+1)(\sqrt{4+x}+2) \\[8pt] &= (5(0)+1)(\sqrt{4+0}+2) \\[8pt] &= (0+1)(2+2) = 4 \end{aligned}
Jawaban D.