the end behavior of f(x)=(x^2 + 3x - 2)/(x^3 - 5x^2 + 4x - 9) using correct limit nota

Explanation:

Step1: Determine degrees of polynomials

The degree of the numerator $n = 2$ (highest - power of $x$ in $x^{2}+3x - 2$) and the degree of the denominator $m=3$ (highest - power of $x$ in $x^{3}-5x^{2}+4x - 9$).

Step2: Use limit rules for rational functions

For a rational function $y=\frac{a_{n}x^{n}+a_{n - 1}x^{n - 1}+\cdots+a_{0}}{b_{m}x^{m}+b_{m - 1}x^{m - 1}+\cdots+b_{0}}$, when $n\lt m$, $\lim_{x\rightarrow\pm\infty}y = 0$. We find $\lim_{x\rightarrow+\infty}\frac{x^{2}+3x - 2}{x^{3}-5x^{2}+4x - 9}$ and $\lim_{x\rightarrow-\infty}\frac{x^{2}+3x - 2}{x^{3}-5x^{2}+4x - 9}$. Divide both the numerator and denominator by $x^{3}$: [ \begin{align*} \lim_{x\rightarrow\pm\infty}\frac{x^{2}+3x - 2}{x^{3}-5x^{2}+4x - 9}&=\lim_{x\rightarrow\pm\infty}\frac{\frac{x^{2}}{x^{3}}+\frac{3x}{x^{3}}-\frac{2}{x^{3}}}{\frac{x^{3}}{x^{3}}-\frac{5x^{2}}{x^{3}}+\frac{4x}{x^{3}}-\frac{9}{x^{3}}}\ &=\lim_{x\rightarrow\pm\infty}\frac{\frac{1}{x}+\frac{3}{x^{2}}-\frac{2}{x^{3}}}{1-\frac{5}{x}+\frac{4}{x^{2}}-\frac{9}{x^{3}}} \end{align*} ] As $x\rightarrow\pm\infty$, $\frac{1}{x}\rightarrow0$, $\frac{3}{x^{2}}\rightarrow0$, $\frac{2}{x^{3}}\rightarrow0$, $\frac{5}{x}\rightarrow0$, $\frac{4}{x^{2}}\rightarrow0$, $\frac{9}{x^{3}}\rightarrow0$. So, $\lim_{x\rightarrow+\infty}\frac{x^{2}+3x - 2}{x^{3}-5x^{2}+4x - 9}=0$ and $\lim_{x\rightarrow-\infty}\frac{x^{2}+3x - 2}{x^{3}-5x^{2}+4x - 9}=0$.

Answer:

$\lim_{x\rightarrow+\infty}\frac{x^{2}+3x - 2}{x^{3}-5x^{2}+4x - 9}=0$, $\lim_{x\rightarrow-\infty}\frac{x^{2}+3x - 2}{x^{3}-5x^{2}+4x - 9}=0$