]> Applications

### Chapter 1: Differentiation; Section 3: Derivatives; page 4

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# Applications

## Rule of L'Hôpital

We already used sequences and functional ratios which are of one of the undefined forms (1.1.3.9), which are copied here for convenience.

 $\left\{\begin{array}{l}\infty -\infty \text{\hspace{0.28em}};\text{ }0\text{\hspace{0.17em}}\infty \text{\hspace{0.28em}};\text{ }\frac{0}{0}\text{\hspace{0.28em}};\text{ }\frac{\infty }{\infty }\\ \text{ }\text{ }{0}^{0}\text{\hspace{0.28em}};\text{ }{1}^{\infty }\text{\hspace{0.28em}};\text{ }{\infty }^{0}\end{array}\right\}$ (1.3.4.1)

The limits of such expressions are usually problems that have no straightforward solutions, and often require experience and ingenuity. The rule of L'Hôpital, based on differentiation, comes to bridge over this difficulty.

The basic rule is

 $\underset{x\to {x}_{0}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to {x}_{0}}{\mathrm{lim}}\frac{\frac{\text{d}f}{\text{d}x}}{\frac{\text{d}g}{\text{d}x}}\text{ }\text{for}\text{ }f\left({x}_{0}\right)=g\left({x}_{0}\right)=0$ (1.3.4.2)

The requirement is that both functions are differentiable in an interval including  x0 . If  x0  is at the edge of a closed interval, the meaning of the limit is one sided. Without entering into mathematical details, the proof's reasoning is as follows:

 $\begin{array}{l}\underset{x\to {x}_{0}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{\Delta x\to 0}{\mathrm{lim}}\frac{f\left({x}_{0}+\Delta x\right)}{g\left({x}_{0}+\Delta x\right)}=\\ =\underset{\Delta x\to 0}{\mathrm{lim}}\frac{f\left({x}_{0}+\Delta x\right)-f\left({x}_{0}\right)}{g\left({x}_{0}+\Delta x\right)-g\left({x}_{0}\right)}=\\ =\underset{\Delta x\to 0}{\mathrm{lim}}\frac{\frac{f\left({x}_{0}+\Delta x\right)-f\left({x}_{0}\right)}{\Delta x}}{\frac{g\left({x}_{0}+\Delta x\right)-g\left({x}_{0}\right)}{\Delta x}}=\underset{x\to {x}_{0}}{\mathrm{lim}}\frac{\frac{\text{d}f}{\text{d}x}}{\frac{\text{d}g}{\text{d}x}}\end{array}\right\}$ (1.3.4.3)

The requirement of  f(x0) = g(x0) = 0  was used above at the second row.

A simple example is

 $\underset{x\to 1}{\mathrm{lim}}\text{\hspace{0.28em}}\frac{\mathrm{ln}x}{1-x}=\underset{x\to 1}{\mathrm{lim}}\text{\hspace{0.28em}}\frac{\frac{1}{x}}{-1}=-1$ (1.3.4.4)

Another example , where one of the derivatives diverges to infinity is

 $\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.28em}}\frac{\mathrm{sin}x}{\sqrt{x}}=\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.28em}}\frac{\mathrm{cos}x}{0.5{x}^{-0.5}}=0$ (1.3.4.5)

One can extend the L'Hôpital's rule by proving that (1.3.4.2) holds also for the cases

• x$\infty$  and for  x→−$\infty$  instead of  xx0
• the limit of ratio is an undefined expression of the form of  $\frac{\infty }{\infty }$  instead of  $\frac{0}{0}$

It may happen that after applying the L'Hopital's rule one obtains again an expression of the form of ratios of zeros or infinities. Then one can apply the L'Hôpital rule again.

Summary

 $\begin{array}{l}\text{For}\text{\hspace{0.28em}}{x}_{0}\text{\hspace{0.28em}}\text{any number}\text{\hspace{0.28em}}\text{or}\text{\hspace{0.28em}}\text{infinity,}\text{\hspace{0.28em}}\text{if}\\ \underset{x\to {x}_{0}}{\mathrm{lim}}\text{\hspace{0.28em}}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to {x}_{0}}{\mathrm{lim}}\text{\hspace{0.28em}}f\left(x\right)}{\underset{x\to {x}_{0}}{\mathrm{lim}}g\left(x\right)}\text{ }\text{yields}\\ \text{expression}\text{\hspace{0.28em}}\text{of}\text{\hspace{0.28em}}\text{type}\text{ }\frac{0}{0}\text{ }\text{or}\text{ }\frac{\infty }{\infty }\\ \text{then}\text{ }\underset{x\to {x}_{0}}{\mathrm{lim}}\text{\hspace{0.28em}}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to {x}_{0}}{\mathrm{lim}}\text{\hspace{0.28em}}\frac{\text{d}f}{\text{d}x}}{\underset{x\to {x}_{0}}{\mathrm{lim}}\text{\hspace{0.28em}}\frac{\text{d}g}{\text{d}x}}\end{array}\right\}$ (1.3.4.6)

If one obtains again an expression of type  $\frac{0}{0}\text{\hspace{0.28em}}$  or  $\frac{\infty }{\infty }$  one can apply the rule on the derivatives etc.

## Combined L'Hôpital's rule

The L' Hôpital's rule covers just part of the undefined forms (1.3.4.1). By the help of few simple tricks they can be used also for the rest.

Case:  $0\text{\hspace{0.17em}}\infty$ .

The solution is formally:

 $0\text{\hspace{0.17em}}\infty =\frac{0}{\frac{1}{\infty }}\text{ }\text{or}\text{ }0\text{\hspace{0.17em}}\infty =\frac{\infty }{\frac{1}{0}}$ (1.3.4.7)

Example:

 $\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}x\mathrm{ln}x=\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\mathrm{ln}x}{{x}^{-1}}=\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{{x}^{-1}}{-{x}^{-2}}=-\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}x=0$ (1.3.4.8)

Case:  $\infty -\infty$ .

Solution:

 $\infty -\infty =\mathrm{ln}\left[\mathrm{exp}\left(\infty -\infty \right)\right]=\mathrm{ln}\frac{\mathrm{exp}\infty }{\mathrm{exp}\infty }$ (1.3.4.9)

Example:

 $\begin{array}{l}\underset{x\to {0}_{+}}{\mathrm{lim}}\left({x}^{-1}+\mathrm{ln}x\right)=\mathrm{ln}\left(\underset{x\to {0}_{+}}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\mathrm{exp}\left({x}^{-1}\right)}{\mathrm{exp}\left(-\mathrm{ln}x\right)}\right)=\\ =\mathrm{ln}\left(\underset{x\to {0}_{+}}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\mathrm{exp}\left({x}^{-1}\right)}{{x}^{-1}}\right)=\mathrm{ln}\left(\underset{x\to {0}_{+}}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{-{x}^{-2}\mathrm{exp}\left({x}^{-1}\right)}{-{x}^{-2}}\right)=\\ =\mathrm{ln}\left(\underset{x\to {0}_{+}}{\mathrm{lim}}\text{\hspace{0.17em}}\mathrm{exp}\left({x}^{-1}\right)\right)=\underset{x\to {0}_{+}}{\mathrm{lim}}\text{\hspace{0.17em}}{x}^{-1}\to \infty \end{array}\right\}$ (1.3.4.10)

Case:  ${0}^{0}$ .

Solution:

 ${0}^{0}=\mathrm{exp}\left[\mathrm{ln}\left({0}^{0}\right)\right]=\mathrm{exp}\left(0\text{\hspace{0.17em}}\mathrm{ln}0\right)=\mathrm{exp}\text{\hspace{0.17em}}\left(\frac{\mathrm{ln}0}{\frac{1}{0}}\right)$ (1.3.4.11)

Example:

 $\begin{array}{l}\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}{\left(\mathrm{sin}x\right)}^{x}=\mathrm{exp}\left[\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}x\mathrm{ln}\left(\mathrm{sin}x\right)\right]=\\ =\mathrm{exp}\left[\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\mathrm{ln}\left(\mathrm{sin}x\right)}{{x}^{-1}}\right]=\mathrm{exp}\left[\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\frac{\mathrm{cos}x}{\mathrm{sin}x}}{-{x}^{-2}}\right]=\\ =\mathrm{exp}\left[\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{-{x}^{2}}{\mathrm{sin}x}\right]=\mathrm{exp}\left[\underset{x\to 0}{\mathrm{lim}}\left(-x\right)\right]=1\end{array}\right\}$ (1.3.4.12)

Case:  1$\infty$ .

Solution:

 ${1}^{\infty }=\mathrm{exp}\left[\mathrm{ln}\left({1}^{\infty }\right)\right]=\mathrm{exp}\left[\infty \text{\hspace{0.17em}}\mathrm{ln}\left(1\right)\right]=\mathrm{exp}\left[\frac{\mathrm{ln}\left(1\right)}{\frac{1}{\infty }}\right]$ (1.3.4.13)

Example:

 $\underset{x\to 0}{\mathrm{lim}}\left[{\left(1+x\right)}^{\frac{1}{x}}\right]=\mathrm{exp}\left[\underset{x\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\mathrm{ln}\left(1+x\right)}{x}\right]=\mathrm{exp}\left[\underset{x\to 0}{\mathrm{lim}}\left(\frac{\frac{1}{1+x}}{1}\right)\right]=e$ (1.3.4.14)

Case:  $\infty$0 .

Solution:

 ${\infty }^{0}=\mathrm{exp}\left[\mathrm{ln}\left({\infty }^{0}\right)\right]=\mathrm{exp}\left[0\text{\hspace{0.17em}}\mathrm{ln}\infty \right]=\mathrm{exp}\left[\frac{\mathrm{ln}\infty }{\frac{1}{0}}\right]$ (1.3.4.15)

Example:

 $\underset{x\to \infty }{\mathrm{lim}}\text{\hspace{0.17em}}\left({x}^{\frac{1}{x}}\right)=\mathrm{exp}\left[\underset{x\to \infty }{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{\mathrm{ln}x}{x}\right)\right]=\mathrm{exp}\left[\underset{x\to \infty }{\mathrm{lim}}\left(\frac{1}{x}\right)\right]=1$ (1.3.4.16)

## Asymptotes

We already know that a function can be represented approximately in the near vicinity of a point by its tangent line. We'll deal now with an approximated representation of a function as a straight line at infinity. We'll start with an example for clarification.

According to (1.2.4.17)   $\underset{x\to ±\infty }{\mathrm{lim}}\left(\mathrm{tanh}x\right)=±1$ . This means that the parallel lines to the  x  axis  y = ±1  are good approximations for the far regions of the function at  x→±$\infty$  accordingly.

Before getting in the mathematical description of an asymptote we'll see graphically this example and a couple more at Fig. tanhx & asymptotes.

We are seeking the representation of the asymptote as a tangent line of a function  f(x)  for  x→±$\infty$ . The asymptote could depend on the sign of the infinity as we saw in the case of  tanhx .

We'll use  $y=kx+b$  as the representation of the asymptote, where  k  is the slope of the line and  b  is the intercept. The slope of the asymptote  k  should correspond to the derivative of the function at the limit of  x→±$\infty$  and the intercept  b  should be the limit of the expression  ykx . In summary:

 $\begin{array}{l}y=kx+b\text{ }\text{is}\text{\hspace{0.28em}}\text{an}\text{\hspace{0.28em}}\text{asymptote}\text{\hspace{0.28em}}\text{of}\text{ }f\left(x\right)\text{ }\text{if}\\ k=\underset{x\to ±\infty }{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\text{d}f}{\text{d}x}\text{ }\text{exists}\text{\hspace{0.28em}}\text{and}\text{\hspace{0.28em}}\text{is}\text{\hspace{0.28em}}\text{finite}\text{ }\text{and}\text{ }\\ b=\underset{x\to ±\infty }{\mathrm{lim}}\left(f\left(x\right)-kx\right)\text{ }\text{exists}\text{\hspace{0.28em}}\text{and}\text{\hspace{0.28em}}\text{is}\text{\hspace{0.28em}}\text{finite}\end{array}\right\}$ (1.3.4.17)

We'll use the examples given at the previous Fig-window.

Example 1.  f(x) = tanhx

 $\begin{array}{l}k=\underset{x\to ±\infty }{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\text{d}\mathrm{tanh}x}{\text{d}x}=\underset{x\to ±\infty }{\mathrm{lim}}\left(1-{\mathrm{tanh}}^{2}x\right)=0\text{ }\\ b=\underset{x\to ±\infty }{\mathrm{lim}}\left(\mathrm{tanh}x\right)=±1\text{}\\ y=kx+b=±1\end{array}\right\}$ (1.3.4.18)

Example 2.  f(x) = tanhx+x

 $\begin{array}{l}k=\underset{x\to ±\infty }{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\text{d}}{\text{d}x}\left(\mathrm{tanh}x+x\right)=1\text{ }\text{ }\\ b=\underset{x\to ±\infty }{\mathrm{lim}}\left(\mathrm{tanh}x+x-x\right)=±1\\ y=kx+b=x±1\end{array}\right\}$ (1.3.4.19)

Example 3.  f(x) = tanhxx

 $\begin{array}{l}k=\underset{x\to ±\infty }{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\text{d}}{\text{d}x}\left(\mathrm{tanh}x-x\right)=-1\text{ }\text{ }\\ b=\underset{x\to ±\infty }{\mathrm{lim}}\left(\mathrm{tanh}x-x+x\right)=±1\\ y=kx+b=-x±1\end{array}\right\}$ (1.3.4.20)

According to (1.3.4.17) the existence of an asymptote requires two limits that should exist and are finite. For example the function  f(x) = √x  has a limit  k = 0  for  x$\infty$  , but  b$\infty$ . This means a line parallel to the  x  axis but situated at  y$\infty$  which is not an asymptote.

Since the asymptote of a function describes the function at infinity well, we can use the asymptote instead of the function at  x→±$\infty$  for finding the slope, meaning that  f(x) = kx+b . The division by  x  yields

 $k=\underset{x\to ±\infty }{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{f\left(x\right)}{x}-\frac{b}{x}\right)=\underset{x\to ±\infty }{\mathrm{lim}}\text{\hspace{0.17em}}\frac{f\left(x\right)}{x}$ (1.3.4.21)

which is equivalent to the  k  obtained by (1.3.4.17). One should notice that in order to obtain the final expression of (1.3.4.21), we used the fact that  b  is a constant (finite). The equivalence does not follow here from the L'Hôpital rule.

The equivalence of the calculation of  k  between (1.3.4.17) and (1.3.4.21) can easily be demonstrated by applying the latter on the three examples above, which is left as a simple exercise for the user.

Example 4. Hyperbola

This example shows how to obtain the asymptote of a function presented in parametric form. The hyperbola in its parametric presentation (1.2.5.41) will be used:

The function is double valued: positive and negative  y , depending on the sign of the parameter  p .

By the use of (1.3.4.21) the slopes of the asymtotes are

$k=\underset{p\to ±\infty }{\mathrm{lim}}\frac{b\mathrm{sinh}p}{a\mathrm{cosh}p}=\frac{b}{a}\underset{p\to ±\infty }{\mathrm{lim}}\left(\mathrm{tanh}p\right)=±\frac{b}{a}$

According to (1.3.4.17), the intercept of the asymptote corresponding to the positive sign of the slope is:

$\left\{\begin{array}{l}\underset{p\to \infty }{\mathrm{lim}}\left(b\mathrm{sinh}p-\frac{b}{a}a\mathrm{cosh}p\right)=b\underset{p\to \infty }{\mathrm{lim}}\left(\mathrm{sinh}p-\mathrm{cosh}p\right)=\\ =b\underset{p\to \infty }{\mathrm{lim}}\left(\frac{\mathrm{exp}p-\mathrm{exp}\left(-p\right)}{2}-\frac{\mathrm{exp}p+\mathrm{exp}\left(-p\right)}{2}\right)=\\ =-b\underset{p\to \infty }{\mathrm{lim}}\mathrm{exp}\left(-p\right)=0\end{array}\right\}$

The same result is obtained for the negative sign of the slope with the limit of $p\to -\infty$.

It is obvious that there is no dependence on the sign of  a , and therefore the result obtained is correct for the whole domain of the function. In summary the asymptotes of the hyperbola are:

 $y=±\frac{b}{a}x$ (1.3.4.22)

This example is illustrated at Fig. Hyperbola-asymptotes .

Example 5. Folium of Descartes.

The parametric form of the Folium of Descartes was used in the exercise 4 of the previous page.

 $\left\{\begin{array}{l}x=\frac{3at}{1+{t}^{3}}\\ y=\frac{3a{t}^{2}}{1+{t}^{3}}\end{array}\right\}$ (1.3.4.23)

The parameter's value  t = 0  corresponds to  (x,y) = (0,0)  and at the limit  $t\to -1$  corresponds to  $x\to -\infty$  and to  $y\to +\infty$ . From (1.3.4.21) one obtains the slope:

$k=\underset{t\to -1}{\mathrm{lim}}\frac{y}{x}=\underset{t\to -1}{\mathrm{lim}}t=-1$

From (1.3.4.17) and by using this slope one obtains the intercept:

$\left\{\begin{array}{l}b=\underset{t\to -1}{\mathrm{lim}}\left(y+x\right)=\underset{t\to -1}{\mathrm{lim}}\left(\frac{3a{t}^{2}+3at}{1+{t}^{3}}\right)=\\ =\underset{t\to -1}{\mathrm{lim}}\left(3at\frac{1+t}{1+{t}^{3}}\right)=-3a\underset{t\to -1}{\mathrm{lim}}\left(\frac{1+t}{1+{t}^{3}}\right)=\\ =-3a\underset{t\to -1}{\mathrm{lim}}\left(\frac{1}{3{t}^{2}}\right)=-a\end{array}\right\}$

Finally the asymptote is:

 $y=-x-a$ (1.3.4.24)

This result together with the solution of exercise 4 from the previous page are used for displaying the Folium of Descartes (1.3.3.1)

${x}^{3}-3axy+{y}^{3}=0$ at Fig. Folium of Descartes

In the examples we used so far, the asymptote was at one side of the function and looking as a bound of the function in the plane. This is not a rule and there are cases where the function oscillates around the asymptote as seen from the following

Example 6. $f\left(x\right)=\frac{\mathrm{sin}x}{x}$

 $\begin{array}{l}k=\underset{x\to ±\infty }{\mathrm{lim}}\frac{f\left(x\right)}{x}=\underset{x\to ±\infty }{\mathrm{lim}}\frac{\mathrm{sin}x}{{x}^{2}}=0\\ b=\underset{x\to ±\infty }{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\mathrm{sin}x}{x}=0\\ y=0\end{array}\right\}$ (1.3.4.25)

Both limits (k  and  b) where obtained by the sandwich rule:  $-\frac{1}{{x}^{2}}\le \frac{\mathrm{sin}x}{{x}^{2}}\le \frac{1}{{x}^{2}}$  and  $-\frac{1}{x}\le \frac{\mathrm{sin}x}{x}\le \frac{1}{x}$ . The example is shown at Fig.  sin(x)/x .

So far we used the presentation of an asymptote as a straight line of the form  y = kx+b . Does it cover everything? No! We left the presentation  x = c , where  c  is a constant, meaning the lines parallel to the y axis.

We have to add to the asymplotes also the lines  x = c  corresponding to the points:

 $\underset{x\to c}{\mathrm{lim}}f\left(x\right)\text{\hspace{0.17em}}\to \text{\hspace{0.17em}}±\infty$ (1.3.4.26)

irrespective of the sign in front of the infinity. As an example  tanx  has such asymptotes at  $x=±\frac{2n-1}{2}\text{\hspace{0.17em}}\pi$ , where  n  is a natural number.

## Exercises

Exercise 1. Calculate the following limit:

Exercise 2. Calculate the following limit:

Exercise 3. Calculate the following limit:

Exercise 4. Find the asymptotes of the hyperbola in its canonical form (without the use of its parametric presentation)

Exercise 5. The following function is given:

1. Find the asymptotes!
2. The function represents a hyperbola. Use the asymptotes of the hyperbola (1.3.4.22) in order to calculate the angle of rotation necessary for transforming it to canonical form!
3. Do the rotation and calculate the corresponding values of a and b of its canonical form:  $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ .

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