Resolu\u00e7\u00e3o<\/b><\/span><\/p>\nOra, $$\\begin{array}{*{20}{l}}
\n{\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\left[ {f(x) – \\left( {3x – 1} \\right)} \\right]}& = &{\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\left[ {\\frac{{k{x^3} – 3{x^2} + x + 1}}{{3{x^2} + 1}} – \\left( {3x – 1} \\right)} \\right]} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{k{x^3} – 3{x^2} + x + 1 – 9{x^3} + 3{x^2} – 3x + 1}}{{3{x^2} + 1}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{\\left( {k – 9} \\right){x^3} – 2x + 2}}{{3{x^2} + 1}}}
\n\\end{array}$$<\/p>\n
Donde $$\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } \\left[ {f(x) – \\left( {3x – 1} \\right)} \\right] = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\left[ {f(x) – \\left( {3x – 1} \\right)} \\right] = 0 \\Leftrightarrow k – 9 = 0 \\Leftrightarrow k = 9$$<\/p>\n
Portanto, para $k=9$, a reta de equa\u00e7\u00e3o $y=3x-1$ \u00e9 ass\u00edntota obl\u00edqua do gr\u00e1fico de $f$, quando ${x \\to\u00a0 – \\infty }$, quer quando ${x \\to\u00a0 + \\infty }$.<\/p>\n
\nAlternativa<\/strong>:<\/p>\n$$\\begin{array}{*{20}{l}}
\n{\\left\\{ {\\begin{array}{*{20}{l}}
\n{\\mathop {\\lim }\\limits_{x \\to _{ + \\infty }^{ – \\infty }} \\frac{{f(x)}}{x} = 3} \\\\
\n{\\mathop {\\lim }\\limits_{x \\to _{ + \\infty }^{ – \\infty }} \\left( {f(x) – 3x} \\right) =\u00a0 – 1}
\n\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{\\mathop {\\lim }\\limits_{x \\to _{ + \\infty }^{ – \\infty }} \\frac{{k{x^3} – 3{x^2} + x + 1}}{{3{x^3} + x}} = 3} \\\\
\n{\\mathop {\\lim }\\limits_{x \\to _{ + \\infty }^{ – \\infty }} \\left( {\\frac{{k{x^3} – 3{x^2} + x + 1}}{{3{x^2} + 1}} – 3x} \\right) =\u00a0 – 1}
\n\\end{array}} \\right.}& \\Leftrightarrow &{} \\\\
\n{}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{\\frac{k}{3} = 3} \\\\
\n{\\mathop {\\lim }\\limits_{x \\to _{ + \\infty }^{ – \\infty }} \\frac{{\\left( {k – 9} \\right){x^3} – 3{x^2} – 2x + 1}}{{3{x^2} + 1}} =\u00a0 – 1}
\n\\end{array}} \\right.}& \\Leftrightarrow &{} \\\\
\n{}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{k = 9} \\\\
\n{k – 9 = 0}
\n\\end{array}} \\right.}& \\Leftrightarrow &{k = 9}
\n\\end{array}$$<\/p>\n<\/p>\n