Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 18

Determine $\displaystyle \lim_{x \to -\infty} \frac{\sqrt{9x^6 - x}}{x^3 + 1}$


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\begin{equation}
\begin{aligned}

\lim_{x \to - \infty} \frac{\sqrt{9x^6 - x}}{x^3 + 1} \cdot \frac{\displaystyle \frac{1}{\sqrt{x^6}}}{\displaystyle \frac{1}{x^3}} =& \lim_{x \to - \infty} \frac{\displaystyle \sqrt{\frac{9 \cancel{x^6}}{\cancel{x^6}}} - \frac{x}{x^6}}{\displaystyle \frac{\cancel{x^3}}{\cancel{x^3}} + \frac{1}{x^3}}
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=& \lim_{x \to - \infty} \frac{\displaystyle \sqrt{9 - \frac{1}{x^5}}}{\displaystyle 1 + \frac{1}{x^3}}
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=& \frac{\displaystyle \lim_{x \to - \infty} \sqrt{9 - \frac{1}{x^5}}}{ \lim \limits_{x \to - \infty} \left( 1 + \frac{1}{x^3} \right) }
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=& \frac{\displaystyle \sqrt{9 - \lim_{x \to - \infty} \frac{1}{x^5}}}{\displaystyle 1 + \lim \limits_{x \to - \infty} \frac{1 }{x^3}}
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=& \frac{\sqrt{9 - 0}}{1 + 0}
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=& \frac{\sqrt{9}}{1}
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=& \frac{3}{1}
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=& 3

\end{aligned}
\end{equation}
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