NZ Level 8 (NZC) Level 3 (NCEA) [In development]

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.

Investigate the behaviour of $f(x)=\frac{3x}{1+x^2}$`f`(`x`)=3`x`1+`x`2 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 $x^2$`x`2 in order to make use of a known limit fact. Thus, $f(x)=\frac{\frac{3}{x}}{\frac{1}{x^2}+1}$`f`(`x`)=3`x`1`x`2+1. Clearly, $f(x)\rightarrow0$`f`(`x`)→0 as $x\rightarrow\pm\infty$`x`→±∞, because $\lim_{x\rightarrow\infty}\frac{1}{x}=0$lim`x`→∞1`x`=0.

We can learn even more without drawing the function. If $x>0$`x`>0, $f(x)=\frac{3x}{1+x^2}$`f`(`x`)=3`x`1+`x`2 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.

Investigate the asymptotic behaviour of $g(x)=\frac{x^3+x+1}{x^2+1}$`g`(`x`)=`x`3+`x`+1`x`2+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`(`x`2+1)+1`x`2+1, we can rewrite the function as $g(x)=x+\frac{1}{x^2+1}$`g`(`x`)=`x`+1`x`2+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.

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`)=`e`−`x`. Since this is just $\frac{1}{e^x}$1`e``x` and $e^x\rightarrow\infty$`e``x`→∞ 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 $e^x$`e``x`, such as $g(x)=\frac{e^x}{\pi-e^x}$`g`(`x`)=`e``x`π−`e``x`.

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`)=1π`e``x`−1. Then, making use of the fact that $\frac{1}{y}\rightarrow0$1`y`→0 as $y\rightarrow\infty$`y`→∞, we see that $g(x)\rightarrow\frac{1}{0-1}=-1$`g`(`x`)→10−1=−1 as $x\rightarrow\infty$`x`→∞.

Note that $g(0)=\frac{1}{\pi-1}$`g`(0)=1π−1 and when $x=\ln\pi$`x`=`l``n`π the function is undefined. Moreover, the closer $x$`x` gets to $\ln\pi$`l``n`π, the larger the function value becomes - values just below $\ln\pi$`l``n`π 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 $e^x\rightarrow0$`e``x`→0 and $\frac{\pi}{e^x}\rightarrow\infty$π`e``x`→∞.

Thus, $g(x)$`g`(`x`) has horizontal asymptotes at $0$0 and $-1$−1 and a vertical asymptote at $x=\ln\pi$`x`=`l``n`π.

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

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 modelled by $P\left(t\right)=\frac{1}{\left(0.997-\frac{t}{29}\right)^{29}}$`P`(`t`)=1(0.997−`t`29)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.

Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DLoading Graph...ALoading Graph...BLoading Graph...CLoading Graph...D

Which function has the properties $\lim_{x\to-4^+}f\left(x\right)=-\infty$lim`x`→−4+`f`(`x`)=−∞ and $\lim_{x\to-4^-}f\left(x\right)=-\infty$lim`x`→−4−`f`(`x`)=−∞?

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

`f`(`x`)=14−`x`A$f\left(x\right)=\frac{x}{\left(x+4\right)^2}$

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

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

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

`f`(`x`)=14−`x`A$f\left(x\right)=\frac{x}{\left(x+4\right)^2}$

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

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

`f`(`x`)=1`x`+4D

Identify discontinuities and limits of functions

Apply differentiation methods in solving problems