Question

Find the limit as x →∞ of {sqroot (9x^2+x)] - 3x}

Answer 1

\[\lim_{x \rightarrow ∞} (\sqrt{9x ^{2}+x}-3x)\]

Answer 2

\[\lim_{x\to\infty}\left(\sqrt{9x^2+x}-3x\right)\cdot\frac{\sqrt{9x^2+x}+3x}{\sqrt{9x^2+x}+3x}\\
\lim_{x\to\infty}\frac{9x^2+x-9x^2}{\sqrt{9x^2+x}+3x}\\
\lim_{x\to\infty}\frac{x}{\sqrt{x^2}\sqrt{9+\frac{1}{x}}+3x}\\
\lim_{x\to\infty}\frac{x}{|x|\sqrt{9+\frac{1}{x}}+3x}\]
\(x\to\infty\) means \(x>0\), so \(|x|=x\):
\[\lim_{x\to\infty}\frac{x}{x\sqrt{9+\frac{1}{x}}+3x}\\
\lim_{x\to\infty}\frac{1}{\sqrt{9+\frac{1}{x}}+3}=\cdots \]

Answer 3

Thank you!