Question

`\displaystyle \lim_(x \to \infty){\frac{7x - 1}{\sqrt[3]{5x^3 + 4x - 1} }}`

Answer 1

If we evaluate at infinity, we get the following:

`\large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_(x \to \infty){\frac{7x - 1}{\sqrt[3]{5x^3 + 4x - 2}}} = (\infty)/(\infty) \end{gathered}$}`

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:

`\large\displaystyle\text{$\begin{gathered}\sf L= \lim_(x \to \infty)\frac{7x-1 }{\sqrt[3]{5x^3+4x-2 } }\cdot((1)/(x) )/((1)/(x) ) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf \ \ = \lim_(x \to \infty)\frac{(7x)/(x)-(1)/(x ) }{\sqrt[3]{(5x^3)/( x^(3) )+(4x)/(x^(3) )-(2)/(x^(3) ) } } \end{gathered}$}\\`

`\large\displaystyle\text{$\begin{gathered}\sf \ \ = \lim_(x \to \infty)\frac{7-(1)/(x) }{\sqrt[3]{ 5+(4 )/(x^(2) ) -(2)/(x^(3) ) } } \end{gathered}$}`

That when "evaluating at infinity" we have:

`\large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_(x \to \infty){\frac{7 - (1)/(x)}{\sqrt[3]{5 + (4)/(x^2) - (2)/(x^3)}}} = \frac{7 - 0}{\sqrt[3]{5 + 0 - 0}} = \frac{7}{\sqrt[3]{5}} \end{gathered}$}`