Determine $k$ de modo que a reta de equação $y = 3x – 1$ seja assíntota do gráfico da função

Ora, $$\begin{array}{*{20}{l}}
{\mathop {\lim }\limits_{x \to  + \infty } \left[ {f(x) – \left( {3x – 1} \right)} \right]}& = &{\mathop {\lim }\limits_{x \to  + \infty } \left[ {\frac{{k{x^3} – 3{x^2} + x + 1}}{{3{x^2} + 1}} – \left( {3x – 1} \right)} \right]} \\
{}& = &{\mathop {\lim }\limits_{x \to  + \infty } \frac{{k{x^3} – 3{x^2} + x + 1 – 9{x^3} + 3{x^2} – 3x + 1}}{{3{x^2} + 1}}} \\
{}& = &{\mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {k – 9} \right){x^3} – 2x + 2}}{{3{x^2} + 1}}}
\end{array}$$

Donde $$\mathop {\lim }\limits_{x \to  – \infty } \left[ {f(x) – \left( {3x – 1} \right)} \right] = \mathop {\lim }\limits_{x \to  + \infty } \left[ {f(x) – \left( {3x – 1} \right)} \right] = 0 \Leftrightarrow k – 9 = 0 \Leftrightarrow k = 9$$

Portanto, para $k=9$, a reta de equação $y=3x-1$ é assíntota oblíqua do gráfico de $f$, quando ${x \to  – \infty }$, quer quando ${x \to  + \infty }$.

Alternativa:

$$\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{\mathop {\lim }\limits_{x \to _{ + \infty }^{ – \infty }} \frac{{f(x)}}{x} = 3} \\
{\mathop {\lim }\limits_{x \to _{ + \infty }^{ – \infty }} \left( {f(x) – 3x} \right) =  – 1}
\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}
{\mathop {\lim }\limits_{x \to _{ + \infty }^{ – \infty }} \frac{{k{x^3} – 3{x^2} + x + 1}}{{3{x^3} + x}} = 3} \\
{\mathop {\lim }\limits_{x \to _{ + \infty }^{ – \infty }} \left( {\frac{{k{x^3} – 3{x^2} + x + 1}}{{3{x^2} + 1}} – 3x} \right) =  – 1}
\end{array}} \right.}& \Leftrightarrow &{} \\
{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}
{\frac{k}{3} = 3} \\
{\mathop {\lim }\limits_{x \to _{ + \infty }^{ – \infty }} \frac{{\left( {k – 9} \right){x^3} – 3{x^2} – 2x + 1}}{{3{x^2} + 1}} =  – 1}
\end{array}} \right.}& \Leftrightarrow &{} \\
{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}
{k = 9} \\
{k – 9 = 0}
\end{array}} \right.}& \Leftrightarrow &{k = 9}
\end{array}$$