Limits problem 05.mws

1. Limits at Infinity.

The precise definition of a limit at infinity is limit(f(x),x = infinity) = L if and only if for every epsilon >0 there is a corresponding N such that abs(f(x)-L) < epsilon whenever N < x .
Consider
limit(sqrt(4*x^2+1)/(x-1),x = infinity) = 2 . Let us illustrate the definition for epsilon = .5e-1 by finding a value of N such that if N < x then abs(sqrt(4*x^2+1)/(x-1)-2) < .5e-1 . To do this, let us plot the function

> f:=x->sqrt(4*x^2+1)/(x-1);

f := proc (x) options operator, arrow; sqrt(4*x^2+1...

over a domain that will reveal the appropriate value of N . Let us first plot the function on the domain 10..200.

> epsilon:=0.05;N:=10;plot([2-epsilon,2+epsilon,f(x)],x=N..200,color=[red,red,blue]);

epsilon := .5e-1

N := 10

[Maple Plot]

What this plot suggests is that the value of the function gets outside of the region we want when x gets to be a little more than 40. Therefore, we should choose N to be about 42. Let us illustrate this fact with a better plot.

> epsilon:=0.05;N:=42;plot([2-epsilon,2+epsilon,f(x)],x=N..200,color=[red,red,blue],axes=boxed);

epsilon := .5e-1

N := 42

[Maple Plot]

It would appear that when epsilon = `0.05` we can take N to be 42.

A good question is how do we really know that as x gets very large (in other words, change the number 200 to a larger number), then the y value will stay within the range 2.05..1.95 ? The answer is that we can not be sure from the plot alone. It is typical that a plot will suggest what the correct answer is, but not allow us to be sure that the answer is correct. One method of verifying the results of the plot is to ask Maple to solve the inequality abs(sqrt(4*x^2+1)/(x-1)-2) < .5e-1 , which it can do for us.

> solve(abs(sqrt(4*x^2+1)/(x-1)-2) < .05,x);

RealRange(Open(41.12158614680527704121436),infinity...

Maple says that in the open interval (41.12158615, infinity ), this inequality will hold. Thus the value N = 42 will satisfy the requirements of the definition.

Submission:

Using the methods outlined above find values for N that correspond to epsilon = .5 and epsilon = .1 for the limit

limit(sqrt(4*x^2+1)/(x+1),x = -infinity) = -2 ( Include a graph like the one above for each of the two values of epsilon . Also notice that in the definition we must replace N < x with x < N , and we should expect N to be negative.)

Submission worksheet:

 

2. More limits at Infinity.

The formal definition of a limit at infinity is not given in th e text, but can be stated as follows.

Suppose that a function f is defined on an open interval including infinity (that is an interval of the form ( a, infinity )). Then we say that the limit of f(x) as is L , and write Limit(f(x) = L,x = infinity) , if for every number epsilon >0 there is a corresponding positive number N , such that abs(f(x)-L) < epsilon whenever N < x .  If Limit(f(x) = L,x = infinity) , then the line y = L is a horizontal asymptote of the curve y = f(x) . There are several ways to explore limits at infinity using Maple. Let us consider the following function.

> f:=x->1+sin(x)/x;

f := proc (x) options operator, arrow; 1+sin(x)/x e...

First, Maple can find limits at infinity by using the limit command.

> limit(f(x),x=infinity);

1

One can even make a plot which gives an idea of the limit at infinity.

> plot(f(x),x=-infinity..infinity,y=0..2);

[Maple Plot]

One disadvantage of this form of display is that it does not show the value of the limit. Another way to get a display of the limit at infinity is to compose the function with the tangent function and note that as , . So if you plot f(tan(x)) , you get a graph that shows the limit as . Also when x->-Pi/2, tan(x)->-infinity, so you can determine limit(f(x),x = -infinity) from the picture as well

> plot(f(tan(x)),x=-Pi/2..Pi/2);

[Maple Plot]

We will use this idea of composing with the tangent function to test the formal definition of the limit at infinity. First, let us use the picture above to guess that limit(f(x),x = infinity) = 1 . Then let us choose epsilon = .1 , and plot the lines y = L-epsilon and y = L+epsilon on the same graph.

> L:=1;epsilon:=.1;

L := 1

epsilon := .1

> plot([f(tan(x)),L-epsilon,L+epsilon],x=0..Pi/2,y=0..2,color=[blue,red,red]);

[Maple Plot]

Now, clicking on the graph, it appears that when x>1.46, the value of the function always lies between these two lines. But since we are plotting the composition of the function and the tangent, this means that when x>tan(1.46), the value of the function lies in the required region. Let us approximate tan(1.46) using maple.

> tan(1.46);

8.988607601724169500744525

We see that if we choose N = 9 in the formal definition, then abs(f(x)-L) < epsilon whenever N < x .

Finally, let us graph the function over a finite interval, along with its asymptote y = 1 . One can see that the function appears to get very close to the asymptote, even though it oscillates above and below it.

> plot([f(x),1], x=0..100);

[Maple Plot]

Submission:

Use the ideas above to study the behavior as and for the following functions:

(a) f(x) = (x^2-2*x+1)/(2*x^2-x+2)

(b) f(x) = (2*x+ln(x))/x

(c) f(x) = cos(1/x)/x

For each function, include a Maple computation of the limit, a graph which allows you to find an N such that abs(f(x)-L) < epsilon whenever N < x , as well as your computation of an N which satisfies the requirement, and a graph which shows the function and its horizontal asymptote as in the same plot.

Submission worksheet: