Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sqrt{x^2+1}-1}{\sqrt{3x^5+4\sin^4 x}} = \cdots \)
- \( 0 \)
- \( \frac{1}{4} \)
- \( \frac{1}{\sqrt{7}} \)
- \( \frac{1}{2} \)
- \( \frac{1}{\sqrt{3}} \)
(SBMPTN 2016)
Pembahasan:
\begin{aligned} \lim_{x \to 0} \ \frac{\sqrt{x^2+1}-1}{\sqrt{3x^5+4\sin^4 x}} &= \lim_{x \to 0} \ \frac{\sqrt{x^2+1}-1}{\sqrt{3x^5+4\sin^4 x}} \cdot \frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1} \\[8pt] &= \lim_{x \to 0} \ \frac{(x^2+1)-1}{(\sqrt{3x^5+4\sin^4 x})(\sqrt{x^2+1}+1)} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} \\[8pt] &= \lim_{x \to 0} \ \frac{x^2 \cdot \frac{1}{x^2}}{\left(\sqrt{ \frac{3x^5}{x^4} + 4 \ \frac{\sin^4 x}{x^4} }\right)(\sqrt{x^2+1}+1)} \\[8pt] &= \frac{1}{\left( \sqrt{3(0) + 4 \cdot (1)^4}\right)(\sqrt{0^2+1}+1)} \\[8pt] &= \frac{1}{(\sqrt{4})(\sqrt{1}+1)} = \frac{1}{(2)(2)} = \frac{1}{4} \end{aligned}
Jawaban B.