# Limit Analysis: Simplifying Rational Expressions

• Shackleford
In summary, it is incorrect to assume that the terms in (2) will all go to 0 as n approaches infinity. This is because the numerator will also increase as n increases. For 4) and 5), the denominators go to 0 while the numerators do not, resulting in a limit of +/- infinity.
Shackleford
1) = 0

The denominator grows much faster than the numerator.

2) = 0

Each of these terms would simply be zero, right?

For 4) and 5), the f(x) = x3

I simplified these to

4) = [(x-2)(x2+2x+4)]/(x-3)

5) = [(x-3)(x2+3x+9)]/(x-2)

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled2.png

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png

Last edited by a moderator:
Shackleford said:
1) = 0

The denominator grows much faster than the numerator.

2) = 0

Each of these terms would simply be zero, right?
(1) is correct, (2) is not. In (2) your sequence is
$$\frac{1}{n}+ \frac{2}{n}+ \cdot\cdot\cdot+ \frac{n}{n^2}= \frac{1+ 2+ 3+ \cdot\cdot\cdot+ n}{n^2}$$

Use the fact that $1+ 2+ 3+ \cdot\cdot\cdot+ n= (1/2)n(n+1)$.

If you are thinking you can take the limit in each term so that you get 0 for each, no that is not correct.

For 4) and 5), the f(x) = x3

I simplified these to

4) = [(x-2)(x2+2x+4)]/(x-3)
You can do that but the obvious point should be that denominator goes to 0 while the numerator goes to f(3)- f(2)= 27- 8 which is NOT 0.

5) = [(x-3)(x2+3x+9)]/(x-2)
Again, the denominator goes to 0 while the numerator does not.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled2.png

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png

Last edited by a moderator:
For 2), I didn't know you could rewrite the sum in that manner.

I then get [(1/2)n(n+1)]/n = (1/2)[(n+1)/n] = 1/2.

If I had to guess, I would say 4) and 5) go to +/- infinity, respectively.

Last edited:
Am I right?

## 1. What is the purpose of finding limits in analysis?

The purpose of finding limits in analysis is to understand the behavior of a function as its input approaches a specific value. It helps to determine the value that a function is approaching or the value that it cannot reach, which is important in many real-world applications.

## 2. How do you find limits algebraically?

To find limits algebraically, you can use various techniques such as factoring, rationalizing, or simplifying the expression. You can also use L'Hopital's rule or the squeeze theorem in some cases. It is important to understand the properties of limits and apply them correctly to find the limit of a given function.

## 3. What is the difference between the left-hand and right-hand limits?

The left-hand limit is the value that a function approaches when the input is approaching from the left side, while the right-hand limit is the value that a function approaches when the input is approaching from the right side. If the left-hand and right-hand limits are equal, then the overall limit exists. If they are not equal, then the limit does not exist.

## 4. How do you use a graph to find limits?

You can use a graph to find limits by visually observing the behavior of the function near the input value. If the function approaches a specific value as the input approaches the given value, then that value is the limit. If the function has a hole or a jump at the given value, then the limit does not exist.

## 5. Can you find limits at infinity?

Yes, you can find limits at infinity by looking at the behavior of the function as the input approaches positive or negative infinity. If the function approaches a finite value, then that value is the limit at infinity. If the function approaches positive or negative infinity, then the limit at infinity does not exist.

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