find the following limit. if the limit is infinite, enter infinity or -infinity. if the limit is not…

Explanation:

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

Divide both the numerator and denominator of $\frac{6 - x^{5}}{x^{3}+6}$ by $x^{3}$. We get $\lim_{x\rightarrow\infty}\frac{\frac{6}{x^{3}}-\frac{x^{5}}{x^{3}}}{\frac{x^{3}}{x^{3}}+\frac{6}{x^{3}}}=\lim_{x\rightarrow\infty}\frac{\frac{6}{x^{3}}-x^{2}}{1 + \frac{6}{x^{3}}}$.

Step2: Evaluate the limit of each term

We know that $\lim_{x\rightarrow\infty}\frac{c}{x^{n}} = 0$ for any constant $c$ and positive integer $n$. So, $\lim_{x\rightarrow\infty}\frac{6}{x^{3}}=0$. Then, $\lim_{x\rightarrow\infty}\frac{\frac{6}{x^{3}}-x^{2}}{1+\frac{6}{x^{3}}}=\frac{\lim_{x\rightarrow\infty}\frac{6}{x^{3}}-\lim_{x\rightarrow\infty}x^{2}}{\lim_{x\rightarrow\infty}1+\lim_{x\rightarrow\infty}\frac{6}{x^{3}}}$. Since $\lim_{x\rightarrow\infty}\frac{6}{x^{3}} = 0$ and $\lim_{x\rightarrow\infty}1=1$, and $\lim_{x\rightarrow\infty}x^{2}=\infty$, the expression becomes $\frac{0-\infty}{1 + 0}$.

Step3: Simplify the result

$\frac{0-\infty}{1+0}=-\infty$.

Answer:

$-\infty$