Answer
$\lim\limits_{k \to \infty} \frac{k}{\sqrt {9k^2+1}} =\frac{1}{3}$
Work Step by Step
$\lim\limits_{k \to \infty} \frac{k}{\sqrt {9k^2+1}} = \lim\limits_{k \to \infty} \frac{\frac{k}{k}}{\frac{\sqrt {9k^2+1}}{k}}= \lim\limits_{k \to \infty} \frac{1}{\sqrt {9+\frac{1}{k^2}}} = \frac{1}{\sqrt {9+0}} = \frac{1}{3}$
