Calculus I - Limits At Infinity, Part II (Practice Problems)

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Section 2.8 : Limits at Infinity, Part II

For problems 1 – 6 evaluate (a) $\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)$ and (b) $\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)$ .

$f\left( x \right) = {{\bf{e}}^{8 + 2x - {x^3}}}$ Solution
$f\left( x \right) = {{\bf{e}}^{\frac{{6{x^2} + x}}{{5 + 3x}}}}$ Solution
$f\left( x \right) = 2{{\bf{e}}^{6x}} - {{\bf{e}}^{ - 7x}} - 10{{\bf{e}}^{4x}}$ Solution
$f\left( x \right) = 3{{\bf{e}}^{ - x}} - 8{{\bf{e}}^{ - 5x}} - {{\bf{e}}^{10x}}$ Solution
$\displaystyle f\left( x \right) = \frac{{{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{8x}}}}{{9{{\bf{e}}^{8x}} - 7{{\bf{e}}^{ - 3x}}}}$ Solution
$\displaystyle f\left( x \right) = \frac{{{{\bf{e}}^{ - 7x}} - 2{{\bf{e}}^{3x}} - {{\bf{e}}^x}}}{{{{\bf{e}}^{ - x}} + 16{{\bf{e}}^{10x}} + 2{{\bf{e}}^{ - 4x}}}}$ Solution

For problems 7 – 12 evaluate the given limit.

$\mathop {\lim }\limits_{t \to \, - \infty } \ln \left( {4 - 9t - {t^3}} \right)$ Solution
$\displaystyle \mathop {\lim }\limits_{z \to \, - \infty } \ln \left( {\frac{{3{z^4} - 8}}{{2 + {z^2}}}} \right)$ Solution
$\displaystyle \mathop {\lim }\limits_{x \to \,\infty } \ln \left( {\frac{{11 + 8x}}{{{x^3} + 7x}}} \right)$ Solution
$\mathop {\lim }\limits_{x \to - \infty } {\tan ^{ - 1}}\left( {7 - x + 3{x^5}} \right)$ Solution
$\displaystyle \mathop {\lim }\limits_{t \to \,\infty } {\tan ^{ - 1}}\left( {\frac{{4 + 7t}}{{2 - t}}} \right)$ Solution
$\displaystyle \mathop {\lim }\limits_{w \to \,\infty } {\tan ^{ - 1}}\left( {\frac{{3{w^2} - 9{w^4}}}{{4w - {w^3}}}} \right)$ Solution