Answer
$$\lim\limits_{x\to\infty}\frac{\sqrt{x+3x^2}}{4x-1}=\frac{\sqrt3}{4}$$
Work Step by Step
$$A=\lim\limits_{x\to\infty}\frac{\sqrt{x+3x^2}}{4x-1}$$$$A=\lim\limits_{x\to\infty}\frac{X}{Y}$$
Divide both numerator and denominator by $x$, we have
- In the numerator:
$X=\frac{\sqrt{x+3x^2}}{x}=\frac{\sqrt{x+3x^2}}{\sqrt {x^2}}=\sqrt{\frac{x+3x^2}{x^2}}=\sqrt{\frac{1}{x}+3}$
- In the denominator:
$Y=\frac{4x-1}{x}=4-\frac{1}{x}$
Therefore, $$A=\lim\limits_{x\to\infty}\frac{\sqrt{\frac{1}{x}+3}}{4-\frac{1}{x}}$$$$A=\frac{\lim\limits_{x\to\infty}\sqrt{\frac{1}{x}+3}}{4-\lim\limits_{x\to\infty}\frac{1}{x}}$$$$A=\frac{\sqrt{\lim\limits_{x\to\infty}(\frac{1}{x})+3}}{4-\lim\limits_{x\to\infty}\frac{1}{x}}$$$$A=\frac{\sqrt{0+3}}{4-0}$$$$A=\frac{\sqrt3}{4}$$
