Section 2.7 : Limits at Infinity, Part I
6. For \(\displaystyle f\left( x \right) = \frac{{{x^3} - 2x + 11}}{{3 - 6{x^5}}}\) answer each of the following questions.
- Evaluate \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\).
- (b) Evaluate \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\).
- Write down the equation(s) of any horizontal asymptotes for the function.
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a Evaluate \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\). Show SolutionTo do this all we need to do is factor out the largest power of \(x\) that is in the denominator from both the denominator and the numerator. Then all we need to do is use basic limit properties along with Fact 1 from this section to evaluate the limit.
\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^3} - 2x + 11}}{{3 - 6{x^5}}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^5}\left( {\frac{1}{{{x^2}}} - \frac{2}{{{x^4}}} + \frac{{11}}{{{x^5}}}} \right)}}{{{x^5}\left( {\frac{3}{{{x^5}}} - 6} \right)}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{\frac{1}{{{x^2}}} - \frac{2}{{{x^4}}} + \frac{{11}}{{{x^5}}}}}{{\frac{3}{{{x^5}}} - 6}} = \frac{0}{{ - 6}} = \require{bbox} \bbox[2pt,border:1px solid black]{0}\]b Evaluate \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\). Show Solution
For this part all of the mathematical manipulations we did in the first part did not depend upon the limit itself and so don’t really need to be redone here. However, it is easy enough to add them in so we’ll go ahead and include them.
\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{{x^3} - 2x + 11}}{{3 - 6{x^5}}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{{x^5}\left( {\frac{1}{{{x^2}}} - \frac{2}{{{x^4}}} + \frac{{11}}{{{x^5}}}} \right)}}{{{x^5}\left( {\frac{3}{{{x^5}}} - 6} \right)}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{\frac{1}{{{x^2}}} - \frac{2}{{{x^4}}} + \frac{{11}}{{{x^5}}}}}{{\frac{3}{{{x^5}}} - 6}} = \frac{0}{{ - 6}} = \require{bbox} \bbox[2pt,border:1px solid black]{0}\]c Write down the equation(s) of any horizontal asymptotes for the function. Show Solution
We know that there will be a horizontal asymptote for \(x \to - \infty \) if \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\) exists and is a finite number. Likewise, we’ll have a horizontal asymptote for \(x \to \infty \) if \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) exists and is a finite number.
Therefore, from the first two parts, we can see that we will get the horizontal asymptote.
\[y = 0\]for both \(x \to - \infty \) and \(x \to \infty \).