Nilai \( \displaystyle \lim_{x \to 5} \ \frac{x^2-25}{\sqrt{x^2+24}-7} = \cdots \)
- 0
- 5
- 7
- 14
- 18
(SPMB 2007)
Pembahasan:
\begin{aligned} \lim_{x \to 5} \ \frac{x^2-25}{\sqrt{x^2+24}-7} &= \lim_{x \to 5} \ \frac{x^2-25}{\sqrt{x^2+24}-7} \times \frac{\sqrt{x^2+24}+7}{\sqrt{x^2+24}+7} \\[8pt] &= \lim_{x \to 5} \ \frac{(x^2-25)(\sqrt{x^2+24}+7)}{(x^2+24)-49} \\[8pt] &= \lim_{x \to 5} \ \frac{(x^2-25)(\sqrt{x^2+24}+7)}{x^2-25} \\[8pt] &= \lim_{x \to 5} \ (\sqrt{x^2+24}+7) \\[8pt] &= \sqrt{5^2 + 24} + 7 = \sqrt{49} + 7 \\[8pt] &= 14 \end{aligned}
Jawaban D.