We often want to study the long-run behavior of a physical process for which a mathematical model has been devised.
As the parameter in the model increases, as time or the variable controlling the process becomes large, we investigate whether the model approaches a limiting state. That is, we look for asymptotic behavior.
Investigate the behavior of $f(x)=\frac{3x}{1+x^2}$f(x)=3x1+x2 as $x$x increases.
Think: As we are not at the moment concerned with the value $x=0$x=0, we can safely divide the numerator and the denominator by $x^2$x2 in order to make use of a known limit fact.
Do:
$f(x)=\frac{\frac{3}{x}}{\frac{1}{x^2}+1}$f(x)=3x1x2+1. Clearly, $f(x)\rightarrow0$f(x)→0 as $x\rightarrow\pm\infty$x→±∞, because $\lim_{x\rightarrow\infty}\frac{1}{x}=0$limx→∞1x=0.
We can learn even more without drawing the function. If $x>0$x>0, $f(x)=\frac{3x}{1+x^2}$f(x)=3x1+x2 is positive, while if $x<0$x<0, $f(x)$f(x) is negative. We also know that $f(0)=0$f(0)=0.
The denominator will never equal zero, so there cannot be a vertical asymptote. Moreover, the function is an odd function, since $f(-x)=-f(x)$f(−x)=−f(x) for all $x$x. This means the graph will have rotational symmetry about the origin. So, the graph of the function $f$f must have a shape something like the following.
Reflect: Does the graph have any horizontal or oblique asymptotes? What is the behavior of the function as the input values decrease?
Investigate the asymptotic behavior of $g(x)=\frac{x^3+x+1}{x^2+1}$g(x)=x3+x+1x2+1.
Think: It seems apparent that since the numerator is a polynomial of higher degree than that in the denominator, the function has no upper or lower bound as $x$x increases. Let's start by rewriting the function to take advantage of our special limits.
Do:
$\lim_{x\to\infty}\frac{x^3+x+1}{x^2+1}$limx→∞x3+x+1x2+1 | $=$= | $\lim_{x\to\infty}\frac{x(x^2+1)+1}{x^2+1}$limx→∞x(x2+1)+1x2+1 |
Rewriting the function |
$=$= | $\lim_{x\to\infty}x+\frac{1}{x^2+1}$limx→∞x+1x2+1 |
Simplifying the fraction |
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$=$= | $\lim_{x\to\infty}x+\lim_{x\to\infty}\frac{1}{x^2+1}$limx→∞x+limx→∞1x2+1 |
Using limit theorems |
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$=$= | $\lim_{x\to\infty}x+0$limx→∞x+0 |
Using $\lim_{x\to\infty}\frac{1}{x}=0$limx→∞1x=0 |
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$\lim_{x\to\infty}\frac{x^3+x+1}{x^2+1}$limx→∞x3+x+1x2+1 | $=$= | $\lim_{x\to\infty}x$limx→∞x |
Simplifying |
We see that the fraction part can be made vanishingly small by allowing $x$x to be large enough in the positive or negative direction. Thus, $g(x)\rightarrow x$g(x)→x as $x\rightarrow\pm\infty$x→±∞.
The line $y=x$y=x is an asymptote for $g(x)$g(x). It has a slope of $1$1.
Reflect: Does the limit of this function exist as the domain values approach infinity? Negative infinity?
Find the asymptotes of $f\left(x\right)=\frac{e^x}{e^3-e^x}$f(x)=exe3−ex.
Write the equations of all asymptotes on the same line, separated by commas.
The population $P$P of stray cats in a town can be modeled by $P\left(t\right)=\frac{1}{\left(0.997-\frac{t}{29}\right)^{29}}$P(t)=1(0.997−t29)29, where $t$t is in months.
The $t$t-value of the vertical asymptote of the function $P(t)$P(t) is called the 'doomsday' value, since the number of stray cats grows infinitely large when $t$t approaches this value.
Find the doomsday value for this town.
Further research showed that this model is appropriate only for the first $24$24 months, and after this point the population growth will slow and begin to taper off. Which of the following graphs best represents a model which incorporates this new information? Choose the most appropriate answer.
Which function has the properties $\lim_{x\to-4^+}f\left(x\right)=-\infty$limx→−4+f(x)=−∞ and $\lim_{x\to-4^-}f\left(x\right)=-\infty$limx→−4−f(x)=−∞?
$f\left(x\right)=\frac{1}{4-x}$f(x)=14−x
$f\left(x\right)=\frac{x}{\left(x+4\right)^2}$f(x)=x(x+4)2
$f\left(x\right)=\frac{1}{\left(x+4\right)^2}$f(x)=1(x+4)2
$f\left(x\right)=\frac{1}{x+4}$f(x)=1x+4