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

Explanation:

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

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

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

As (x\rightarrow\infty), (\frac{9}{x^{2}}\rightarrow0), (\frac{7}{x^{3}}\rightarrow0), (\frac{2}{x^{3}}\rightarrow0), (\frac{6}{x^{2}}\rightarrow0) 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})

Answer:

(8)