Section 2.8 : Limits at Infinity, Part II
2. For \(f\left( x \right) = {{\bf{e}}^{\frac{{6{x^2} + x}}{{5 + 3x}}}}\) 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 SolutionFirst notice that,
\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{6{x^2} + x}}{{5 + 3x}} = - \infty \]If you aren’t sure about this limit you should go back to the previous section and work some of the examples there to make sure that you can do these kinds of limits.
Now, recalling Example 1 from this section, we know that because the exponent goes to negative infinity in the limit the answer is,
\[\mathop {\lim }\limits_{x \to \, - \infty } {{\bf{e}}^{\frac{{6{x^2} + x}}{{5 + 3x}}}} = \require{bbox} \bbox[2pt,border:1px solid black]{0}\]b \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) Show Solution
First notice that,
\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{6{x^2} + x}}{{5 + 3x}} = \infty \]If you aren’t sure about this limit you should go back to the previous section and work some of the examples there to make sure that you can do these kinds of limits.
Now, recalling Example 1 from this section, we know that because the exponent goes to infinity in the limit the answer is,
\[\mathop {\lim }\limits_{x \to \,\infty } {{\bf{e}}^{\frac{{6{x^2} + x}}{{5 + 3x}}}} = \require{bbox} \bbox[2pt,border:1px solid black]{\infty }\]