Respuesta :
If we evaluate at infinity, we get the following:
[tex]\large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_{x \to \infty}{\frac{7x - 1}{\sqrt[3]{5x^3 + 4x - 2}}} = \frac{\infty}{\infty} \end{gathered}$}[/tex]
Which is an indeterminacy of ∞ / ∞. Therefore, we must multiply and divide by the highest degree monomial (either in numerator or denominator). In this case it is 1/x—since the denominator involves a cube root, then the "monomial" x³ is considered to be a monomial of degree 1—. So:
[tex]\large\displaystyle\text{$\begin{gathered}\sf L= \lim_{x \to \infty}\frac{7x-1 }{\sqrt[3]{5x^3+4x-2 } }\cdot\frac{\frac{1}{x} }{\frac{1}{x} } \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf \ \ = \lim_{x \to \infty}\frac{\frac{7x}{x}-\frac{1}{x } }{\sqrt[3]{\frac{5x^3}{ x^{3} }+\frac{4x}{x^{3} }-\frac{2}{x^{3} } } } \end{gathered}$}\\[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf \ \ = \lim_{x \to \infty}\frac{7-\frac{1}{x} }{\sqrt[3]{ 5+\frac{4 }{x^{2} } -\frac{2}{x^{3} } } } \end{gathered}$}[/tex]
That when "evaluating at infinity" we have:
[tex]\large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_{x \to \infty}{\frac{7 - \frac{1}{x}}{\sqrt[3]{5 + \frac{4}{x^2} - \frac{2}{x^3}}}} = \frac{7 - 0}{\sqrt[3]{5 + 0 - 0}} = \frac{7}{\sqrt[3]{5}} \end{gathered}$}[/tex]
