Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 8

Find the limit $\displaystyle \lim_{x \to \infty} \sqrt{\frac{12x^3 - 5x + 2}{1 + 4x^2 + 3x^3}}$ and justify each step by indicating the appropriate properties of limits.


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\begin{equation}
\begin{aligned}

\lim_{x \to \infty} \sqrt{\frac{12x^3 - 5x + 2}{1 + 4x^2 + 3x^3}} \cdot \frac{\displaystyle \frac{1}{\sqrt{x^3}}}{\frac{1}{\sqrt{x^3}}}
=& \lim_{x \to \infty} \sqrt{\frac{\displaystyle \frac{1}{x^3} (12x^3 - 5x + 2)}{\displaystyle \frac{1}{x^3} (1 + 4x^2 + 3x^3)}}
&&
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=& \lim_{x \to \infty} \sqrt{\frac{\displaystyle \frac{12 \cancel{x^3}}{\cancel{x^3}} - \frac{5x}{x^3} + \frac{2}{x^3}}{\displaystyle \frac{1}{x^3} + \frac{4x^2}{x^3} + \frac{3 \cancel{x^3}}{\cancel{x^3}}}}
&&
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=& \lim_{x \to \infty} \sqrt{\frac{\displaystyle 12 - \frac{5}{x^2} + \frac{2}{x^3}}{\displaystyle \frac{1}{x^3} + \frac{4}{x} + 3}}
&& \text{Apply} \displaystyle \lim_{x \to a} [\sqrt[n]{f(x)}] = \sqrt[n]{\lim_{x \to a} f(x)}
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=& \sqrt{\lim_{x \to \infty} \frac{\displaystyle 12 - \frac{5}{x^2} + \frac{2}{x^3} }{\displaystyle \frac{1}{x^3} + \frac{4}{x} + 3 }}
&& \text{Apply $\large \lim \limits_{x \to a} \left[ \frac{f(x)}{g(x)} \right] = \frac{\lim \limits_{x \to a} f(x)}{\lim_{x \to a} g(x)}$}
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=& \sqrt{ \frac{\displaystyle \lim_{x \to \infty} \left( 12 - \frac{5}{x^2} + \frac{2}{x^3} \right) }{\displaystyle \lim_{x \to \infty} \left( \frac{1}{x^3} + \frac{4}{x} + 3 \right)} }
&& \text{Apply $\lim \limits_{x \to a} [f(x) \pm g(x)] = \displaystyle \lim \limits_{x \to a} f(x) \pm \lim \limits_{x \to a} g(x)$}
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=& \sqrt{\frac{\displaystyle 12 - \lim_{x \to \infty} \frac{5}{x^2} + \lim_{x \to \infty} \frac{z}{x^3} }{\displaystyle \lim_{x \to \infty} \frac{1}{x^3} + \lim_{x \to \infty} \frac{4}{x} + 3}}
&& \text{Apply $\large \lim \limits_{x \to \infty} \displaystyle \frac{1}{x^n} = 0$}
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=& \sqrt{\frac{12 - 0 + 0}{0 + 0 + 3}}
&&
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=& \sqrt{\frac{12}{3}} = \sqrt{4} = 2
&&

\end{aligned}
\end{equation}
$