Question

$\lim_{x \to \infty} \sqrt[2]{9x^{2} +x } -3x$

Answer 1

Temos:

$\sqrt{9x^{2} + x} - 3x = \\\\(\sqrt{9x^{2} + x} - 3x).\frac{(\sqrt{9x^{2} + x} + 3x)}{(\sqrt{9x^{2} + x} + 3x)} = \\\\\frac{9x^2 + x - 9x^2}{\sqrt{9x^{2} + x} + 3x} = \\\\\frac{x}{\sqrt{9x^{2} + x} + 3x} = \\\\\frac{x}{\sqrt{9x^{2} + x} + 3x}.\frac{1/x}{1/x} = \\\\\frac{x.(1/x)}{(1/x).(\sqrt{9x^2 + x} + 3x)} =\\\\\frac{x.(1/x)}{(1/x).\sqrt{9x^2 + x} + (1/x).3x)} = \\\\\frac{1}{\sqrt{\frac{9x^2 + x}{x^2}} + 3} =\\\\\frac{1}{\sqrt{9 + 1/x}+ 3}$

Logo:

$\lim_{x \to \infty} \sqrt{9x^2 + x} - 3x = \lim_{x \to \infty} \frac{1}{\sqrt{9 + 1/x}+ 3} =\frac{1}{\sqrt{9} + 3} = \frac{1}{6}$