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