Question

13-14Evaluate the limit and justify each step by indicating the appropriate properties of limits.

14. \(\mathop {{\bf{lim}}}\limits_{x \to \infty } \,\,\sqrt {\frac{{{\bf{9}}{x^{\bf{3}}} + {\bf{8}}x - {\bf{4}}}}{{{\bf{3}} - {\bf{5}}x + {x^{\bf{3}}}}}} \)

Short answer

The value of the given expression is 3.

Step-by-step solution

Step 1:Simplify the limit using the properties

Divide numerator and denominator of the given expression by \({x^3}\) (Inside the radical).

\(\mathop {\lim }\limits_{x \to \infty } \sqrt {\frac{{9{x^3} + 8x - 4}}{{3 - 5x + {x^3}}}} = \mathop {\lim }\limits_{x \to \infty } \sqrt {\frac{{\frac{{9{x^3} + 8x - 4}}{{{x^3}}}}}{{\frac{{3 - 5x + {x^3}}}{{{x^3}}}}}} \)

Simplify the limit using the properties

Use Law 5 of the properties of limit.

\(\mathop {\lim }\limits_{x \to \infty } \sqrt {\frac{{\frac{{9{x^3} + 8x - 4}}{{{x^3}}}}}{{\frac{{3 - 5x + {x^3}}}{{{x^3}}}}}} = \sqrt {\frac{{\mathop {\lim }\limits_{x \to \infty } \left( {9 + \frac{8}{{{x^2}}} - \frac{4}{{{x^3}}}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\frac{3}{{{x^3}}} - \frac{5}{{{x^2}}} + 1} \right)}}} \)

Simplify the limit using the properties

Use Law 1 and law 2 of the properties of limit.

\(\sqrt {\frac{{\mathop {\lim }\limits_{x \to \infty } \left( {9 + \frac{8}{{{x^2}}} - \frac{4}{{{x^3}}}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\frac{3}{{{x^3}}} - \frac{5}{{{x^2}}} + 1} \right)}}} = \sqrt {\frac{{\mathop {\lim }\limits_{x \to \infty } 9 + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{8}{{{x^2}}}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{4}{{{x^3}}}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\frac{3}{{{x^3}}}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{5}{{{x^2}}}} \right) + \mathop {\lim }\limits_{x \to \infty } 1}}} \)

Simplify the limit using the properties

Use law 8 and law 3, the properties of limit.

\(\begin{array}{c}\sqrt {\frac{{\mathop {\lim }\limits_{x \to \infty } 9 + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{8}{{{x^2}}}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{4}{{{x^3}}}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\frac{3}{{{x^3}}}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{5}{{{x^2}}}} \right) + \mathop {\lim }\limits_{x \to \infty } 1}}} = \sqrt {\frac{{9 + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{8}{{{x^2}}}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{4}{{{x^3}}}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\frac{3}{{{x^3}}}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{5}{{{x^2}}}} \right) + 1}}} \\ = \sqrt {\frac{{9 + 8\left( 0 \right) - 4\left( 0 \right)}}{{3\left( 0 \right) - 5\left( 0 \right) + 1}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\rm{Using}}\,{\rm{theorem 5}}} \right)\\ = \sqrt 9 \\ = 3\end{array}\)

So, the value of the limit is 3.