find the following infinite limit.\n lim_{x\rightarrow+infty}\frac{(x - 2)^2(x - 3)}{(x^2 - 4)(x^2 - 6x +…

Explanation:

Step1: Divide numerator and denominator by highest - power of x

When dealing with limits at infinity of rational functions, we divide both the numerator and denominator by the highest - power of x in the denominator. Here, the highest power of x in the denominator of $\lim_{x\rightarrow\infty}\frac{2x^{2}+8x - 3}{2 - 3x^{2}}$ is $x^{2}$. So we have $\lim_{x\rightarrow\infty}\frac{\frac{2x^{2}}{x^{2}}+\frac{8x}{x^{2}}-\frac{3}{x^{2}}}{\frac{2}{x^{2}}-\frac{3x^{2}}{x^{2}}}=\lim_{x\rightarrow\infty}\frac{2+\frac{8}{x}-\frac{3}{x^{2}}}{\frac{2}{x^{2}}-3}$.

Step2: Use the limit rules for $\frac{1}{x}$ as $x\rightarrow\infty$

We know that $\lim_{x\rightarrow\infty}\frac{1}{x}=0$ and $\lim_{x\rightarrow\infty}\frac{1}{x^{n}} = 0$ for $n>0$. Applying these rules, we get $\frac{\lim_{x\rightarrow\infty}2+\lim_{x\rightarrow\infty}\frac{8}{x}-\lim_{x\rightarrow\infty}\frac{3}{x^{2}}}{\lim_{x\rightarrow\infty}\frac{2}{x^{2}}-\lim_{x\rightarrow\infty}3}$. Since $\lim_{x\rightarrow\infty}2 = 2$, $\lim_{x\rightarrow\infty}\frac{8}{x}=0$, $\lim_{x\rightarrow\infty}\frac{3}{x^{2}}=0$, $\lim_{x\rightarrow\infty}\frac{2}{x^{2}}=0$, and $\lim_{x\rightarrow\infty}3 = 3$, the limit becomes $\frac{2 + 0-0}{0 - 3}$.

Answer:

$-\frac{2}{3}$