Answer
$$\lim\limits_{x\to\infty}(\sqrt{9x^2+x}-3x)=\frac{1}{6}$$
Work Step by Step
$$A=\lim\limits_{x\to\infty}(\sqrt{9x^2+x}-3x)$$$$A=\lim\limits_{x\to\infty}\Bigg[(\sqrt{9x^2+x}-3x)\times\frac{\sqrt{9x^2+x}+3x}{\sqrt{9x^2+x}+3x}\Bigg]$$$$A=\lim\limits_{x\to\infty}\frac{(9x^2+x)-9x^2}{\sqrt{9x^2+x}+3x}$$$$A=\lim\limits_{x\to\infty}\frac{x}{\sqrt{9x^2+x}+3x}$$$$A=\lim\limits_{x\to\infty}\frac{X}{Y}$$
Divide both numerator and denominator by $x$, we have
- In the numerator:
$X=\frac{x}{x}=1$
- In the denominator:
$Y=\frac{\sqrt{9x^2+x}+3x}{x}=\frac{\sqrt{9x^2+x}}{x}+\frac{3x}{x}=\sqrt{\frac{9x^2+x}{x^2}}+3=\sqrt{9+\frac{1}{x}}+3$
Therefore, $$A=\lim\limits_{x\to\infty}\frac{1}{\sqrt{9+\frac{1}{x}}+3}$$$$A=\frac{1}{\lim\limits_{x\to\infty}(\sqrt{9+\frac{1}{x}})+3}$$$$A=\frac{1}{\sqrt{9+\lim\limits_{x\to\infty}\frac{1}{x}}+3}$$$$A=\frac{1}{\sqrt{9+0}+3}$$$$A=\frac{1}{6}$$
