Answer
$$\lim\limits_{x\to\infty}\frac{x^4-3x^2+x}{x^3-x+2}=\infty$$
Work Step by Step
$$A=\lim\limits_{x\to\infty}\frac{x^4-3x^2+x}{x^3-x+2}$$$$A=\lim\limits_{x\to\infty}\frac{X}{Y}$$
Divide both numerator and denominator by $x^3$ (the highest-power variable of the denominator), we have
- The numerator:
$X=\frac{x^4-3x^2+x}{x^3}=x-\frac{3}{x}+\frac{1}{x^2}$
- The denominator:
$Y=\frac{x^3-x+2}{x^3}=1-\frac{1}{x^2}+\frac{2}{x^3}$
Therefore, $$A=\lim\limits_{x\to\infty}\frac{x-\frac{3}{x}+\frac{1}{x^2}}{1-\frac{1}{x^2}+\frac{2}{x^3}}$$
As $x\to\infty$, we have $x-\frac{3}{x}+\frac{1}{x^2}\to(\infty-3\times0+0)=\infty$
while $1-\frac{1}{x^2}+\frac{2}{x^3}\to(1-0+2\times0)=1$
Therefore, $$A=\infty$$
