Question

find the limit. (if the limit is infinite, enter \\( \infty \\) or \\( -\infty \\), as appropriar
\\( \lim _ { x \
ightarrow - \infty } \frac { \sqrt { 1 + 16 x ^ { 6 } } } { 5 - x ^ { 3 } } \\)

Explanation:

Step1: Analyze the highest - degree terms

When \(x\to-\infty\), for the numerator \(\sqrt{1 + 16x^{6}}\), since \(x^{6}\geq0\) for all real \(x\), and when \(|x|\) is large, \(\sqrt{1+16x^{6}}\approx\sqrt{16x^{6}}\). Because \(x\to-\infty\), \(\sqrt{16x^{6}}=- 4x^{3}\) (since \(\sqrt{x^{6}}=|x^{3}|=-x^{3}\) when \(x\lt0\)). The denominator \(5 - x^{3}\approx - x^{3}\) when \(x\to-\infty\).

Step2: Calculate the limit

$$
LATEXBLOCK0
$$

Answer:

\(4\)