evaluate the limit using the appropriate properties of limits. (if the limit is infinite, er\n lim _{x…

Explanation:

Step1: Divide numerator and denominator by (x^{3})

$$ \begin{align*} \lim_{x\rightarrow\infty}\sqrt{\frac{64x^{3}+9x - 7}{2-6x+x^{3}}}&=\lim_{x\rightarrow\infty}\sqrt{\frac{\frac{64x^{3}}{x^{3}}+\frac{9x}{x^{3}}-\frac{7}{x^{3}}}{\frac{2}{x^{3}}-\frac{6x}{x^{3}}+\frac{x^{3}}{x^{3}}}}\ &=\lim_{x\rightarrow\infty}\sqrt{\frac{64+\frac{9}{x^{2}}-\frac{7}{x^{3}}}{\frac{2}{x^{3}}-\frac{6}{x^{2}} + 1}} \end{align*} $$

Step2: Apply the limit property (\lim_{x\rightarrow\infty}\frac{1}{x^{n}} = 0) ((n>0))

As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{9}{x^{2}}=0), (\lim_{x\rightarrow\infty}\frac{7}{x^{3}}=0), (\lim_{x\rightarrow\infty}\frac{2}{x^{3}}=0), (\lim_{x\rightarrow\infty}\frac{6}{x^{2}}=0)

So, (\lim_{x\rightarrow\infty}\sqrt{\frac{64+\frac{9}{x^{2}}-\frac{7}{x^{3}}}{\frac{2}{x^{3}}-\frac{6}{x^{2}}+1}}=\sqrt{\frac{64 + 0-0}{0 - 0+1}})

Step3: Simplify the expression

(\sqrt{\frac{64}{1}}=\sqrt{64}=8)

Answer:

(8)