Answer
$0.17$
Work Step by Step
Step 1. See table, we can set a range of x-values and calculate the function values to estimate the limit as $0.166$
Step 2. See graph, we can confirm the result graphically.
Step 3. Algebraically, $\lim_{x\to\infty}(\sqrt {9x^2+x}-3x)=\lim_{x\to\infty}\frac{(\sqrt {9x^2+x}-3x)(\sqrt {9x^2+x}+3x)}{\sqrt {9x^2+x}+3x}=\lim_{x\to\infty}\frac{9x^2+x-9x^2}{\sqrt {9x^2+x}+3x}=\lim_{x\to\infty}\frac{1}{\sqrt {9+1/x}+3}=\frac{1}{\sqrt {9}+3}=\frac{1}{6}$
