Question

Question on Infinite Limit, involving a Radical. lim x→∞√(4x^2+3x) −2x

Answer 1

\[\lim_{x \rightarrow \infty} \sqrt{4x ^{2} + 3x} -2x\]

Answer 2

\[\lim_{x\to\infty}\left(\sqrt{4x^2+3x}-2x\right)\]
\[\sqrt{4x^2+3x}-2x\cdot\frac{\sqrt{4x^2+3x}+2x}{\sqrt{4x^2+3x}+2x}\\
\frac{4x^2+3x-4x^2}{\sqrt{4x^2+3x}+2x}\\
\frac{3x}{\sqrt{4x^2+3x}+2x}\\
\frac{3x}{\sqrt{x^2}\sqrt{4+\frac{3}{x}}+2x}\\
\frac{3x}{|x|\sqrt{4+\frac{3}{x}}+2x}\]
Since \(x\to\infty\), you have \(|x|=x\):
\[\frac{3x}{x\sqrt{4+\frac{3}{x}}+2x}\\
\frac{3}{\sqrt{4+\frac{3}{x}}+2}\]
Taking the limit is easy from here.

Answer 3

Thank you.