I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 2.8 : Limits at Infinity, Part II
5. For \(\displaystyle f\left( x \right) = \frac{{{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{8x}}}}{{9{{\bf{e}}^{8x}} - 7{{\bf{e}}^{ - 3x}}}}\) evaluate each of the following limits.
- \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\)
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a \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\) Show SolutionThe exponential with the negative exponent is the only term in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the negative exponent in the denominator from both the numerator and denominator to evaluate this limit.
\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{8x}}}}{{9{{\bf{e}}^{8x}} - 7{{\bf{e}}^{ - 3x}}}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{{\bf{e}}^{ - 3x}}\left( {1 - 2{{\bf{e}}^{11x}}} \right)}}{{{{\bf{e}}^{ - 3x}}\left( {9{{\bf{e}}^{11x}} - 7} \right)}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{1 - 2{{\bf{e}}^{11x}}}}{{9{{\bf{e}}^{11x}} - 7}} = \frac{{1 - 0}}{{0 - 7}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{7}}}\]b \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) Show Solution
The exponential with the positive exponent is the only term in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the positive exponent in the denominator from both the numerator and denominator to evaluate this limit.
\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{8x}}}}{{9{{\bf{e}}^{8x}} - 7{{\bf{e}}^{ - 3x}}}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{{{\bf{e}}^{8x}}\left( {{{\bf{e}}^{ - 11x}} - 2} \right)}}{{{{\bf{e}}^{8x}}\left( {9 - 7{{\bf{e}}^{ - 11x}}} \right)}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{{{\bf{e}}^{ - 11x}} - 2}}{{9 - 7{{\bf{e}}^{ - 11x}}}} = \frac{{0 - 2}}{{9 - 0}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{2}{9}}}\]