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
4. For \(f\left( x \right) = 3{{\bf{e}}^{ - x}} - 8{{\bf{e}}^{ - 5x}} - {{\bf{e}}^{10x}}\) 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 SolutionHere we have two exponents with negative exponents and so both will go to infinity in the limit. However, each term has opposite signs and so each term seems to be suggesting different answers for the limit.
In order to determine which “wins out” so to speak all we need to do is factor out the term with the most negative exponent and then use basic limit properties.
\[\mathop {\lim }\limits_{x \to \, - \infty } \left( {3{{\bf{e}}^{ - x}} - 8{{\bf{e}}^{ - 5x}} - {{\bf{e}}^{10x}}} \right) = \mathop {\lim }\limits_{x \to \, - \infty } \left[ {{{\bf{e}}^{ - 5x}}\left( {3{{\bf{e}}^{4x}} - 8 - {{\bf{e}}^{15x}}} \right)} \right] = \left( \infty \right)\left( { - 8} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty }}\]b \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) Show Solution
For this limit the exponentials with negative exponents will simply go to zero and there is only one exponential with a positive exponent (which will go to infinity) and so there isn’t much to do with this limit.
\[\mathop {\lim }\limits_{x \to \,\infty } \left( {3{{\bf{e}}^{ - x}} - 8{{\bf{e}}^{ - 5x}} - {{\bf{e}}^{10x}}} \right) = 0 - 0 - \infty = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty }}\]