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# Infinite and Non-Existent Limits

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Infinite and Non-Existent Limits

You have heard time and again that it is "against the rules" to divide by the number 0 . Even the most basic calculator will return some form of "ERROR" if you try to divide even the smallest of numbers by 0 .

Do you really understand why it is "against the rules"? What is really wrong with dividing by nothing?

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### Guidance

Infinite limits

Functions can exhibit a number of different behaviors as the input value gets very large or very small.

As $x$ approaches $\infty$ , some functions output values closer and closer to a single number, some approach zero, and some continue to get larger and larger or smaller and smaller without limit.

In this lesson, we will explore functions of the last type, functions with infinite limits , and the different types of asymptotes they may have.

#### Example A

Evaluate the function $h(x)=\frac{x^2}{2x-1}$ .

Solution

To evaluate this function, consider the behavior of the function as larger and larger values are Inserted for x . As x approaches $\infty$ , the function values also approach $\infty$ . Therefore the limit of the function as x approaches $\infty$ is: $\lim_{x \to \infty} \frac{x^2} {2x - 1} = \infty$ . Similarly, as x approaches $-\infty$ , f ( x ) approaches $-\infty$ . Therefore we have $\lim_{x \to -\infty} \frac{x^2} {2x - 1} = -\infty.$

We can also understand this limit if we analyze the equation for h ( x ). As x gets larger and larger, the value of the expression 2 x - 1 gets closer and closer to the value of the expression 2 x . That is, for sufficiently large values of x , 2 x - 1 $\approx$ 2 x . Therefore the values of h ( x ) approach $\frac{x^2}{2x}=\frac{x}{2}$ . As x gets larger and larger, so does $\frac{x}{2}$ . For large values of x , the function h ( x ) gets closer and closer to $\frac{x}{2}$ . Therefore the limit is infinity.

#### Example B

Approximate the function f ( x ) = x 2 + 2 x - 3.

Solution

This function has an infinite limit as x approaches infinity. However, as x gets larger and larger, f ( x ) $\approx$ x 2 , since the x 2 value grows much more quickly than the 2 x value, particularly apparent at very large +/- values of x . If this is not immediately apparent, evaluate the function for x = 1,000,000, and you will quickly get the idea!

Therefore we can use the function y = x 2 to describe the end behavior of f ( x ).

#### Example C

Describe the end behavior of each graph. That is, determine if the function has a limit L , if the limit is infinite, or if the limit does not exist.

a.) $\,\! y=x^2$
b.) $\,\! y=2(-1)^x$
c.) $y=1 - \frac{1} {|x|}$

Solution:

a.) $\,\! y=x^2$

As x approaches $+\infty$ , x 2 also approaches $+\infty$ . As x approaches $-\infty$ , x 2 approaches $+\infty$ . Therefore $\lim_{x \to \infty} x^2=\lim_{x \to -\infty} x^2=\infty$ .

b.) $\,\! y=2(-1)^x$

This function is difficult to understand without producing a graph. The table shows that the function only takes on two values: 2, and -2, and is undefined at non-integer values of x . As x approaches $+\infty$ or $-\infty$ , the function values alternate between 2 and -2. Therefore the limit does not exist.

c.) $y=1 - \frac{1} {|x|}$

If you look at the table of this function, which has negative and positive values of x , you can see that as x approaches $+\infty$ or $-\infty$ , the function values approach 1.

Therefore $\lim_{x \to \infty} \left (1 - \frac{1} {|x|}\right)=\lim_{x \to -\infty} \left (1 - \frac{1} {|x|}\right)=1$ .

We can also determine this limit analytically. For large values of x , | x | is also large, and so $\frac{1}{|x|}$ is small (since dividing 1 by a large number results in a very small number). Therefore, for large values of x , $1 - \frac{1} {|x|} \approx 1 - 0 = 1$ . We can make the same argument for x approaching $-\infty$ .

Concept question wrap-up

Dividing by zero is "against the rules" because there is no definition for the answer you would get.

Consider what happens as you take a given value and divide it by smaller and smaller numbers:

2 / 10 = 1/5
2 / 1 = 2
2 / .1 = 20
2 / .001 = 2,000
2 / .000000001 = 2,000,000,000

As we divide by progressively smaller numbers, the quotient gets larger and larger. Also, we can see that in each case, the problem could be reversed by multiplying the product by the dividend to get the divisor, for instance: 2 / .1 = 20 ==> 20 * .1 = 2.

Unfortunately, this doesn't work if you actually divide by 0, even if you assume that dividing by zero resulted in infinity! No matter how big the number you multiply by zero, even infinity, you will never be able to get back to 2.

$\therefore x/0 = undefined$

### Vocabulary

An infinite function is one whose output approaches infinity or negative infinity as very large or very small values are calculated for the input variable (usually " $x$ ").

An asymptote is a line representing a value toward which the value of a function may approach, but never actually reach.

### Guided Practice

Questions

1) Evaluate $\lim_{x \to \infty} \frac{x^3}{x-7}$
2) Evaluate $\lim_{x \to \infty} \frac{x^2 +2x}{x - 3}$
3) Evaluate $\lim_{x \to \infty} x^3$
4) Describe the end behavior of $y = 2x^2 + 3x - 7$
5) Evaluate $-2x^3 - 5x^2 + 8x$

Solutions

1) The degree of the numerator is greater than the degree of the denominator, so the function will grow without bound. Since the denominator is x - 7 , the function cannot include x = 7 , because the function cannot be defined where the denominator = 0 . Logically, as x gets huge, the -7 matters less and less, and we end up with just x 2 .
2) Similar to the last problem, the numerator is of greater degree than the denominator, so the function does not approach a limit. The denominator is x - 3 , so the graph cannot include 3 .
3) As $x \to \infty$ , or in other words, "as x gets huge", the value of x 3 grows even faster, either positive or negative, so there is no limit.
4) As x grows huge, x 2 grows much faster than the rest of the expression, therefore, we can approximate the end behavior of $2x^2 + 3x - 7$ with $y = x^2$
5) $-2x^3 - 5x^2 + 8x$ is a 3rd degree equation, so it will turn twice, since it is not a rational function, there are no concerns about numerator or denominator. The function will have no limits, and will grow without bound in both the positive and negative directions. If you use a graphing calculator to graph the function, you will see that $y = x^3$ can be used to approximate it.

### Practice

Problems 1-3: The limit of $f(x)$ as $x \to t$ cannot exist if which conditions are true? List three conditions.

1. Give an example of a limit that does not exist

Problems 5-7: Assuming that $f(x)$ is a rational function:

1. What is $\lim_{x \to \infty} f(x)$ when the degree of the numerator is less than the degree of the denominator?
2. What is $\lim_{x \to \infty} f(x)$ when the degrees of the numerator and the denominator are equal?
3. What is $\lim_{x \to \infty} f(x)$ when the degree of the numerator is greater than the degree of the denominator?
1. In general, if r is a positive real number, what is $\lim_{x \to \infty}\frac{1}{x^r}$ ?
2. In general, if r is a positive real number, what is $\lim_{x \to \infty} x^r$ ?

Problems 10-13: Let a and b be real numbers and let t be a positive integer. Complete each of the following properties of limits.

1. $\lim_{x \to a} x^t =$
2. If f is a polynomial, $\lim_{x \to a} f(x) =$
3. $\lim_{x \to a} k \cdot f(x) =$
4. $\lim_{x \to \infty} t^x =$

Problems 14-16: Evaluate.

1. $\lim_{x \to -5}\frac{(5 + x)^2 - 25}{x}$
2. $\lim_{x \to 3}\frac{x^3 - 6x + 2}{x^2 + 2x - 3}$
3. $\lim_{x \to 0}\frac{(3 + 3y)^{-1} - 3y^{-1}}{x}$