Asymptotes
- Page ID
- 217661
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Asymptotes Horizontal Asymptotes We define a horizontal asymptote of a function as the limit as \(x\) approaches infinity (or negative infinity). Symbolically, we write
We can also take the limit as \( x \) approaches negative infinity and also call the result a horizontal asymptote of \(f(x)\). For rational functions the limits are always the same. On the other hand absolute value and root functions can have two different horizontal asymptotes.
Example Find the horizontal asymptote of \( \displaystyle \lim_{x \to \infty} \frac{3x^2 - x + 2}{2x^2 + 4x + 5} \)
Solution We divide numerator and denominator by the highest power of \(x \) (\( x^2\)). \( \displaystyle \lim_{x \to \infty} \frac{3 - \frac{1}{x} + \frac{2}{x^2}}{2 + \frac{4}{x} + \frac{5}{x^2}} \)
Now when we plug in, we get 3/2. That is 3/2 is a horizontal asymptote. The graph is shown below.
Exercises Find the horizontal asymptotes of the following. Hold your mouse on the yellow rectangle for the answer. A. \( y = \frac{3x^3 - 5x + 1}{6x^3 + 3x^2 - 4} \) B. \( y = \frac{7x^2 - 3x + 2}{5x^3 - 4x^2 + 10} \)
Example Find the horizontal asymptote of \( y = \frac{4x + 1}{\sqrt{x^2 - 9}} \)
Solution We find the limit \( \displaystyle \lim_{x \to \infty} \frac{4x + 1}{\sqrt{x^2 - 9}} \) by dividing numerator and denominator by \( x \). \( \displaystyle = \lim_{x \to \infty} \frac{4 + \frac{1}{x}}{\frac{1}{x} \sqrt{(x^2 - 9)}} \) \( \displaystyle = \lim_{x \to \infty} \frac{4 + \frac{1}{x}}{\sqrt{\frac{1}{x^2}{(x^2 - 9)}}} \) \( \displaystyle = \lim_{x \to \infty} \frac{4 + \frac{1}{x}}{\sqrt{1 - \frac{9}{x^2}}} = \frac{4}{\sqrt{1}} = 4\) We are not done, since when we encounter an irrational function, we must check the left horizontal asymptote. We next take the limit as \( x \) approaches negative infinity. Again we divide both numerator and denominator by \( x \). We get
\( \displaystyle = \lim_{x \to -\infty} \frac{4 + \frac{1}{x}}{\frac{1}{x} \sqrt{(x^2 - 9)}} \)
We can not bring the \( \frac{1}{x} \) into the square root sign since it is negative. For \( x \) negative, \( \frac{1}{x} = -\sqrt{\frac{1}{x^2}}\) Substituting gives \( \displaystyle = \lim_{x \to -\infty} \frac{4 + \frac{1}{x}}{\frac{1}{x} \sqrt{(x^2 - 9)}} \) \( \displaystyle = \lim_{x \to -\infty} \frac{ 4 + \frac{1}{x} }{ -\sqrt{ \frac{1}{x^2} }\sqrt{(x^2 - 9)}}\) \( \displaystyle = \lim_{x \to -\infty} \frac{4 + \frac{1}{x}}{-\sqrt{1 - \frac{9}{x^2}}} = \frac{-4}{\sqrt{1}} = -4\) We can conclude that the function has a left horizontal asymptote of -4 and a right horizontal asymptote of 4. The graph is shown below.
Vertical Asymptotes
A reduced rational function will have a vertical asymptote when the denominator is 0. Once we know that a function has a vertical asymptote at \(x = c\), we then need to find the limit as \( x \) approaches \( c \) form the left and from the right. The limit will always be infinity or negative infinity, so we only need to check for the sign of the limit.
Example: Describe the vertical asymptote of \( y = \frac{1}{x - 1}
Solution We see that when \( x = 1\), the function will have a vertical asymptote. note that to the left of 1 (plug in 0.9) the function is negative and to the right of 1 (plug in 1.1) the function is positive, hence \( \displaystyle \lim_{x \to 1^{-}} f(x) = -\infty \) and \( \displaystyle \lim_{x \to 1^{+}} f(x) = \infty \)
Application
Example: If a person jumps from an airplane, his velocity is modeled by \( v(t) = \frac{-32t}{0.01t + 1}\) What is his terminal velocity?
Solution Terminal velocity means the velocity after an infinite amount of time. We find \( \displaystyle \lim_{t \to \infty} \frac{-32t}{0.01t + 1} = \frac{-32}{0.01 + \frac{1}{t}} = \frac{-32}{0.01} = -3200 \) The terminal velocity of the jumper is 3200 feet per second downward.
Example The population \(P\) after \( t\) years of a newly introduced species of wildcat can be modeled by the equation \( P(t) = \frac{900t + 8000}{3t + 4000}\) What is the equilibrium population (the population at time approaches infinity)?
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