Lab Overview

Why are we taking time to develop l'Hopital's rule? Part of the answer is that the method does not make sense until you know about derivatives. Another part of the reason is that the indeterminate forms that can be evaluated with l'Hopital's rule are going to appear in several contexts within the next month or so of this course. The limits encoutnered in this lab have been selected specifically for their importance later in this course.

Example 1:  (p. 407, # 4)

Let's start with a problem that is not at all obvious.

>	F := arctan(3*x)/arcsin(x): L := Limit( F, x=0 ): L;

When you first look at this problem, and note that , it is difficult to know where to start. Most people just do not have the familiarity with the inverse trigonometric functions to be able to "see" the value of this limit. To begin to understand the problem, plot the function  on an interval containing the origin. Because the arcsin function is defined only on , that is the largest domain on which this function is defined.

>	p1 := plot( F, x=-1..-0.01, y=0..5 ): p2 := plot( F, x=0.01..1, y=0..5 ): display( [p1,p2] );

>

The plot suggests that the value of the limit will be 3. Let's see if we can confirm this using properties of limits. Because the denominator converges to 0 as x->0 there is a hope that l'Hopital's rule will apply. To check this, obtain the numerator and denominator and check their limits as ->0:

>	f := numer(F): Lf := Limit( f, x=0 ): Lf = value( Lf );

>	g := denom(F): Lg := Limit( g, x=0 ): Lg = value( Lg );

Because this limit has form , l'Hopital's rule can be applied. The derivatives of the numerator and denominator are

>	f1 := Diff( f, x ): g1 := Diff( g, x ): f1 = value( f1 ); g1 = value( g1 );

so

>	F1 := value( f1/g1 ): f1/g1 = F1;

The original limit can be replaced by

>	L1 := Limit( F1, x=0 ): L1;

This limit is easy to evaluate:

>	L1 = eval( F1, x=0 );

Thus, as predicted from the plot,

>	L = L1; `` = eval( F1, x=0 );

To conclude, let's look at the plot of the two functions considered in this problem.

>	p3 := plot( F1, x=-1..1, y=0..5, color=green ): display( [p1,p2,p3] );

>

Note that the functions are different, but do have the same limit at .

>

Example 2:  

The function

>	F := x*ln(x): `F(x)`=F;

is defined only for  > 0. A plot of this function on an interval to the right of the origin

>	p1 := plot( F, x=0.01..3 ): p1;

suggests that

>	L := Limit( F, x=0, right ): L = 0;

Now, let's see if we can confirm this. The original limit is has indeterminate form . To apply l'Hopital's rule it is necessary to convert this to a  or  form. This can be done by writing the original problem as either

   =        (  )

or

   =          (  )

The first form looks more appealing because it is easier to differentiate  than . To verify that l'Hopital's rule applies

>	f := ln(x): Lf := Limit( f, x=0, right ): Lf = value( Lf );

>	g := 1/x: Lg := Limit( g, x=0, right ): Lg = value( Lg );

Applying l'Hopital's rule means computing derivatives

>	f1 := Diff( f, x ): g1 := Diff( g, x ): f1 = value( f1 ); g1 = value( g1 );

so

>	F1 := value( f1/g1 ): f1/g1 = F1;

The original limit can be replaced by

>	L1 := Limit( F1, x=0, right ): L1;

This limit is easy to evaluate:

>	L1 = eval( F1, x=0 );

Thus, as predicted from the plot,

>	L = L1; `` = eval( F1, x=0 );

To conclude, let's look at the plot of both functions considered in this problem.

>	p2 := plot( F1, x=0..1, color=green ): display( [p1,p2] );

>

Note that the functions are different, but do have the same limit at .

>

Example 3:  (  > 0)

The limit variable, , is assumed to be an integer. In this form, l'Hopital's rule is not applicable. The usual approach to this problem is to convert the problem to an equivalent problem involving a real variable. In most situations this is done by simply replacing all occurrences of  with :

>	Ln := Limit( a^(1/n), n=infinity ): Lx := Limit( a^(1/x), x=infinity ): Ln = Lx;

To emphasize that these problems are equivalent -- and to gain insight into the value of this limit -- consider the following plot. This plot contains the

>

G := (a,n) -> a^(1/n): p1 := plot( [seq([n,G(5,n)],n=1..10 )], style=point, color=blue,             view=[0..10,0..5], legend=["a=5, discrete points"] ): p2 := plot( G(5,x), x=1..10, color=cyan,             legend=["a=5, continuous variable"] ): p3 := plot( [seq([n,G(1/5,n)],n=1..10 )], style=point, color=red,             view=[0..10,0..5], legend=["a=1/5, discrete points"] ): p4 := plot( G(1/5,x), x=1..10, color=pink,             legend=["a=1/5, continuous variable"] ): display([p1,p2,p3,p4]);

>

Evaluation of this integral, for any  > 0, begins by rewriting

   =  =  

Then,

   =  = .

The limit that interests us now is

>	L := Limit( ln(a)/x, x=infinity ): L;

In this limit, the denominator has a limit of  while the numerator has a finite limit that can be either positive (  > 1), negative (0 <  < 1 ) or zero (  = 1 ). This is not an indeterminate form; l'Hopital's rule cannot be applied!  However, it is easily seen that

   =  =  = 0

To conclude,

   =  =  = 1.

This is consistent with the plot.

>

Example 4:  

This problem has an indeterminate form of . L'Hopital's rule can be used only after it has been rewritten with an indeterminate form of  or . Prior to rewriting the problem, however, let's see if a plot can suggest if this limit exists and what its value might be.

>	plot( (1+x)^(1/x), x=0.01..10, y=0..3 );

This picure suggests the limit exists and has a value around 2.7.

To rewrite this problem in a form suitable for l'Hopital's rule, note

   =  = .

This means

In the new limit

>	F := ln(1+x)/x: L := Limit( F, x=0, right ): L;

the fact that the numerator and denominator satisfy

>	f := numer( F ): Lf := Limit( f, x=0, right ): Lf = value( Lf );

>	g := denom( F ): Lg := Limit( g, x=0, right ): Lg = value( Lg );

means this limit has indeterminate form 0/0. An application of l'Hopital's rule leads to

>	f1 := Diff( f, x ): g1 := Diff( g, x ): f1 = value( f1 ); g1 = value( g1 );

so

>	F1 := value( f1/g1 ): f1/g1 = F1;

The original limit can be replaced by

>	L1 := Limit( F1, x=0, right ): L1;

This limit is easy to evaluate:

>	L1 = eval( F1, x=0 );

Thus,

>	L = L1; `` = eval( F1, x=0 );

and so, the original limit is

   =  =  =  

To conclude, note that

>	exp(1) = evalf(exp(1));

which is consistent with the conjecture made at the beginning of this problem.

>

Example 5:  

>	F := x^(5/2) * exp(-x/30): P1 := plot( F, x=0..10 ): P1;

Based on this picture, it looks like the function might increase forever. But, 10 is not a very big number.

>	P2 := plot( F, x=0..100 ): P2;

Now it is clear that something more happens but we are not yet in a position to say anything about the limit at .

>	P3 := plot( F, x=0..1000 ): P3;

This is a little more convincing - but how do we know nothing different happens for  > 1000?

Example 6: p. 413, # 48

>	unassign('n');

>	F := (n,x) -> n^2*x*exp(-n*x): f[n](x) = F(n,x);

>

Lab Questions

1. p. 407, #   8

2. p. 408, # 32

3. p. 413, #   2

4  

5  

6. [Essay Question] Explain the steps needed to show that

    for all real values of  >= 0 and all real values of  > 0

Include how you rewrite the limit in an indeterminate form suitable to apply l'Hopital's rule and how many times l'Hopital's rule is applied. (Remember that  and  are real real numbers, not integers.) Also, explain how this limit justifies the following statement:

 every exponentially growing function grows faster than every nonnegative power for large values of the variable.

>