NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Applications of Limits
Lesson

We often want to study the long-run behaviour 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 behaviour.

Example 1

Investigate the behaviour of  $f(x)=\frac{3x}{1+x^2}$f(x)=3x1+x2 as $x$x increases.

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. Thus, $f(x)=\frac{\frac{3}{x}}{\frac{1}{x^2}+1}$f(x)=3x1x2+1. Clearly, $f(x)\rightarrow0$f(x)0 as $x\rightarrow\pm\infty$x±, because $\lim_{x\rightarrow\infty}\frac{1}{x}=0$limx1x=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 have $f(0)=0$f(0)=0.  

The denominator is never 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.

 

Example 2

Investigate the asymptotic behaviour of $g(x)=\frac{x^3+x+1}{x^2+1}$g(x)=x3+x+1x2+1.

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. By the division algorithm or by noticing that $g(x)=\frac{x(x^2+1)+1}{x^2+1}$g(x)=x(x2+1)+1x2+1, we can rewrite the function as $g(x)=x+\frac{1}{x^2+1}$g(x)=x+1x2+1.

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 $h(x)=x$h(x)=x is an asymptote for $g(x)$g(x). It has a gradient of $1$1.

 

Example 3

Expressions involving exponentials can lead to functions that exhibit asymptotic behaviour. A simple example is the function defined on the real numbers, given by $f(x)=e^{-x}$f(x)=ex. Since this is just $\frac{1}{e^x}$1ex and $e^x\rightarrow\infty$ex as $x\rightarrow\infty$x, it follows that $f(x)\rightarrow0$f(x)0 as $x\rightarrow\infty$x. That is, the function has a horizontal asymptote.

We might investigate more complicated expressions involving $e^x$ex, such as $g(x)=\frac{e^x}{\pi-e^x}$g(x)=exπex.

We note that for very large $x$x, the numerator and denominator are almost the same, except for sign. We can rewrite the expression as $g(x)=\frac{1}{\frac{\pi}{e^x}-1}$g(x)=1πex1. Then, making use of the fact that $\frac{1}{y}\rightarrow0$1y0 as $y\rightarrow\infty$y, we see that $g(x)\rightarrow\frac{1}{0-1}=-1$g(x)101=1 as $x\rightarrow\infty$x.

Note that $g(0)=\frac{1}{\pi-1}$g(0)=1π1 and when $x=\ln\pi$x=lnπ the function is undefined. Moreover, the closer $x$x gets to $\ln\pi$lnπ, the larger the function value becomes - values just below $\ln\pi$lnπ give very large, negative function values, and values just above give very large, positive function values.

When $x\rightarrow-\infty$x, $g(x)\rightarrow0$g(x)0 because $e^x\rightarrow0$ex0 and $\frac{\pi}{e^x}\rightarrow\infty$πex.

Thus, $g(x)$g(x) has horizontal asymptotes at $0$0 and $-1$1 and a vertical asymptote at $x=\ln\pi$x=lnπ.

Example 4

Find the asymptotes of $f\left(x\right)=\frac{e^x}{e^3-e^x}$f(x)=exe3ex.

Write the equations of all asymptotes on the same line, separated by commas.

Example 5

The population $P$P of stray cats in a town can be modelled by $P\left(t\right)=\frac{1}{\left(0.997-\frac{t}{29}\right)^{29}}$P(t)=1(0.997t29)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.

  1. Find the doomsday value for this town.

  2. 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.

    Loading Graph...

    A

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    B

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    C

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    D

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    A

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    B

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    C

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    D

Example 6

Which function has the properties $\lim_{x\to-4^+}f\left(x\right)=-\infty$limx4+f(x)= and $\lim_{x\to-4^-}f\left(x\right)=-\infty$limx4f(x)=?

  1. $f\left(x\right)=\frac{1}{4-x}$f(x)=14x

    A

    $f\left(x\right)=\frac{x}{\left(x+4\right)^2}$f(x)=x(x+4)2

    B

    $f\left(x\right)=\frac{1}{\left(x+4\right)^2}$f(x)=1(x+4)2

    C

    $f\left(x\right)=\frac{1}{x+4}$f(x)=1x+4

    D

    $f\left(x\right)=\frac{1}{4-x}$f(x)=14x

    A

    $f\left(x\right)=\frac{x}{\left(x+4\right)^2}$f(x)=x(x+4)2

    B

    $f\left(x\right)=\frac{1}{\left(x+4\right)^2}$f(x)=1(x+4)2

    C

    $f\left(x\right)=\frac{1}{x+4}$f(x)=1x+4

    D

Outcomes

M8-10

Identify discontinuities and limits of functions

91578

Apply differentiation methods in solving problems

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