Section 2.8 : Limits at Infinity, Part II
For problems 1 – 11 evaluate (a)\(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\) and (b) \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\).
- \(f\left( x \right) = {{\bf{e}}^{{x^4} + 8x}}\)
- \(f\left( x \right) = {{\bf{e}}^{2x + 4{x^2} + 2{x^5}}}\)
- \(\displaystyle f\left( x \right) = {{\bf{e}}^{\frac{{3 - {x^3}}}{{x + {x^2}}}}}\)
- \(\displaystyle f\left( x \right) = {{\bf{e}}^{\frac{{5 - 9x}}{{7 + 3x}}}}\)
- \(\displaystyle f\left( x \right) = {{\bf{e}}^{\frac{{5 + 2{x^6}}}{{x - 8{x^4}}}}}\)
- \(f\left( x \right) = {{\bf{e}}^x} + 12{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{ - 10x}}\)
- \(f\left( x \right) = 9{{\bf{e}}^{2x}} - 7{{\bf{e}}^{ - 14x}} - {{\bf{e}}^x}\)
- \(f\left( x \right) = 20{{\bf{e}}^{ - 8x}} - {{\bf{e}}^{5x}} + 3{{\bf{e}}^{2x}} - {{\bf{e}}^{ - 7x}}\)
- \(\displaystyle f\left( x \right) = \frac{{6{{\bf{e}}^{4x}} + {{\bf{e}}^{ - 15x}}}}{{11{{\bf{e}}^{4x}} + 6{{\bf{e}}^{ - 15x}}}}\)
- \(\displaystyle f\left( x \right) = \frac{{{{\bf{e}}^{3x}} + 9{{\bf{e}}^{ - x}} - 4{{\bf{e}}^{10x}}}}{{2{{\bf{e}}^{7x}} - {{\bf{e}}^{ - x}}}}\)
- \(\displaystyle f\left( x \right) = \frac{{3{{\bf{e}}^{ - 14x}} - {{\bf{e}}^{18x}}}}{{{{\bf{e}}^{ - x}} - 2{{\bf{e}}^{20x}} - {{\bf{e}}^{ - 9x}}}}\)
For problems 12 – 20 evaluate the given limit.
- \(\mathop {\lim }\limits_{x \to \,\infty } \ln \left( {5{x^2} + 12x - 6} \right)\)
- \(\mathop {\lim }\limits_{y \to \, - \infty } \ln \left( {5 - 7{y^5}} \right)\)
- \(\displaystyle \mathop {\lim }\limits_{x \to \,\infty } \ln \left( {\frac{{3 + x}}{{1 + 5{x^3}}}} \right)\)
- \(\displaystyle \mathop {\lim }\limits_{t \to \, - \infty } \ln \left( {\frac{{2t - 5{t^3}}}{{4 + 3{t^2}}}} \right)\)
- \(\displaystyle \mathop {\lim }\limits_{z \to \,\infty } \ln \left( {\frac{{10z + 8{z^2}}}{{{z^2} - 1}}} \right)\)
- \(\mathop {\lim }\limits_{x \to - \infty } {\tan ^{ - 1}}\left( {7 + 4x - {x^3}} \right)\)
- \(\mathop {\lim }\limits_{w \to \infty } {\tan ^{ - 1}}\left( {4{w^2} - {w^6}} \right)\)
- \(\displaystyle \mathop {\lim }\limits_{t \to \,\infty } {\tan ^{ - 1}}\left( {\frac{{4{t^3} + {t^2}}}{{1 + 3t}}} \right)\)
- \(\displaystyle \mathop {\lim }\limits_{z \to \, - \,\infty } {\tan ^{ - 1}}\left( {\frac{{{z^4} + 4}}{{3{z^2} + 5{z^3}}}} \right)\)