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### 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.

1. $$\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)$$
2. $$\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)$$
Hint : Remember that you’ll need to factor the term in the denominator that is causing the denominator to go to infinity from both the numerator and denominator in order to evaluate this limit.

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a $$\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)$$ Show Solution

The 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.

$\mathop {\lim }\limits_{x \to \,\infty } \frac{{{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{8x}}}}{{9{{\bf{e}}^{8x}} - 7{{\bf{e}}^{ - 3x}}}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{{{\bf{e}}^{8x}}\left( {{{\bf{e}}^{ - 11x}} - 2} \right)}}{{{{\bf{e}}^{8x}}\left( {9 - 7{{\bf{e}}^{ - 11x}}} \right)}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{{{\bf{e}}^{ - 11x}} - 2}}{{9 - 7{{\bf{e}}^{ - 11x}}}} = \frac{{0 - 2}}{{9 - 0}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{2}{9}}}$