Wednesday, May 2, 2007

Precalculus 12.4 Limits at Infinity and Limits of Sequences

12.4 Limits at Infinity and Limits of Sequences

Definition of Limits at Infinity
If "f" is a function and L1 and L2 are real numbers, the statements

lim f(x) = L1
x®-¥

AND

lim f(x) = L1
x®+¥

denote the limits at infinity. The first is read "the limit of f(x) as x approaches negative infinity is L1," and the second is read "the limit of f(x) as x approaches infinity is L2"

To help evaluate limits at infinity, you can use the following:

Limits at Infinity
If "r" is a positive real number, then

lim (1/(xr)) = 0 Limit toward the right.
x®+¥

Furthermore, if xr is defined when x is less then 0, then

lim (1/(xr) = 0 the limit toward the left.
x®- ¥

Thought: -1/10 = -.1, -1/100 = -.01, -1/1000 = -.001, -1/10000 = -.0001
so as you can see, in the denominator as x approaches negative infinity, f(x) approaches zero.

Example 1: Evaluating a Limit at Infinity:

lim (5/(2x)) = 0
x®¥

As you can see when you graph this function, as x gets larger, f(x) gets closer to zero so that is how we can conclude the the limit equals zero.

Example 2:
lim (1 - 3/(x2)) =
x®¥

We can separate each part of the function to:

lim (1 )= 1
x®¥

and

lim (3/(x2)) = 0
x®¥

so
lim (1 - 3/(x2)) = 1 - 0 = 1
x®¥

Compare the following limits:
Find the limit as x approaches infinity:

a). f(x) = (-5x + 2)/(4x2 - 1)

b). g(x) = (4x2 + 1)/(3x2 - 1)

c). j(x) = (-3x4 + 1)/(2x3-1)

When you take the limit of each of these functions as x approaches infinity, you get:

1st, look in the denominator and find the largest exponent and use this term by dividing all the terms of the function by that variable raised to the largest exponent.

a).
lim (-5x + 2)/(4x2 - 1)=
x®¥

so divide each term by x2

lim [(-5x)/(x2) + 2/(x2)]/[(4x2)/(x2) - 1/(x2)]
x®¥

= (-0 + 0)/(4 - 0) = 0

b).
lim (4x2 + 1)/(3x2 - 1)=
x®¥

so divide each term by x2

lim [4 + 1/(x2)]/[3 - 1/(x2)]=
x®¥

(4 + 0)/(3 - 0) = 4/3

c).
lim (-3x4 + 1)/(2x3-1)
x®¥

and so divide each of the terms by x3

lim (-3x + 1/x3)/(2 - 1/x3)=
x®¥

-3x/2 so by putting infinity in for x, the limit does not exist because the numerator decreases without bound as the denominator approaches 2.

Therefore we can conclude:

Limits at Infinity for Rational Functions

for the rational function f(x) = N(x)/D(x) when

N(x) = anxn + ... + a0

And

D(x) = bmxm + ... + b0

the limit as x approaches positive or negative infinity is as follows:

lim f(x) = 0, when n is less than m
x®¥

lim f(x) = an/bm, when n = m
x®¥

And if n is greater than m, the limit does not exist.

Limit of a Sequence:
Let "f" be a function of a real variable, such that
lim f(x) = L
x®¥

If {an} is a sequence such that f(n) = an
for every positive integer "n", then the
lim an = L
n®¥

Example:
Given: 1/3, 1/9, 1/27, 1/81, ...

As n increases without bound, the terms of the sequence
an = 1/3n is said to

CONVERGE to zero.

lim 1/3n = 0
n®¥

___________________________________________

A sequence that does not converge is said to DIVERGE!

Example of this is: the sequence 1, -1, 1, -1, ... this diverges.

Find the limit of the Sequence:

Example 1:
lim (4x - 1)/(x + 3)=
x®¥
{3/4, 7/5, 11/6, 15/7, ...} = 4

Example 2:
lim (3x + 2)/(2x2 - 1)=
x®¥
{5/1, 8/7, 11/17, 14/31, ...} = 0

Example 3:
lim (7x2 - 1)/(8x2)=
x®¥
{6/8, 27/32, 62/72, 111/128, ...} = 7/8

Use the site: http://www.calculus-help.com/funstuff/phobe.html

Chapter 1, lesson 4 to help with this lesson.