find the limit. (if the limit is infinite, enter ∞ or -∞, as appropriate. if the limit does not otherwise…

Explanation:

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

Since ( x\to\infty), we know that ( x^3=\sqrt{x^6}). [ \begin{align*} \lim_{x\to\infty}\frac{x^3}{\sqrt{x^6 + 7}}&=\lim_{x\to\infty}\frac{x^3/x^3}{\sqrt{x^6+7}/x^3}\ &=\lim_{x\to\infty}\frac{1}{\sqrt{\frac{x^6 + 7}{x^6}}}\ \end{align*} ]

Step2: Simplify the expression inside the square - root

Using the property ( \frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}), we have (\frac{x^6+7}{x^6}=1+\frac{7}{x^6}). [ \begin{align*} \lim_{x\to\infty}\frac{1}{\sqrt{\frac{x^6 + 7}{x^6}}}&=\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{7}{x^6}}}\ \end{align*} ]

Step3: Use the limit property (\lim_{x\to\infty}\frac{1}{x^n}=0) for (n>0)

As (x\to\infty), (\lim_{x\to\infty}\frac{7}{x^6}=0). Then (\lim_{x\to\infty}\sqrt{1+\frac{7}{x^6}}=\sqrt{1 + 0}=1). [ \begin{align*} \lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{7}{x^6}}}&=\frac{1}{1}\ \end{align*} ]

Answer:

(1)