Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 16

Determine $\displaystyle \lim_{x \to \infty} \frac{x + 2}{\sqrt{9x^2 + 1}}$


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\begin{equation}
\begin{aligned}

\lim_{x \to \infty} \frac{x + 2}{\sqrt{9x^2 + 1}} \cdot \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{\sqrt{x^2}}} =& \lim_{x \to \infty} \frac{\displaystyle \frac{\cancel{x}}{\cancel{x}} + \frac{2}{x}}{\displaystyle \sqrt{\frac{9 \cancel{x^2}}{\cancel{x^2}} + \frac{1}{x^2}}}
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=& \lim_{x \to \infty} \frac{\displaystyle 1 + \frac{2}{x}}{\displaystyle \sqrt{9 + \frac{1}{x^2}}}
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=& \frac{\displaystyle \lim_{x \to \infty} \left(1 + \frac{2}{x} \right) }{\displaystyle \lim_{x \to \infty} \sqrt{9 + \frac{1}{x^2}}}
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=& \frac{\displaystyle 1 + \lim_{x \to \infty} \frac{2}{x}}{\displaystyle \sqrt{9 + \lim_{x \to \infty} \frac{4}{x^2}}}
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=& \frac{1 + 0}{\sqrt{9 + 0}}
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=& \frac{1}{\sqrt{9}}
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=& \frac{1}{3}

\end{aligned}
\end{equation}
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