# Limit of Rational Expression Above or Below

• Jun 11th 2009, 04:25 PM
Yale
Limit of Rational Expression Above or Below
How can you tell if the limit is from above or below?
• Jun 11th 2009, 04:48 PM
VonNemo19
Quote:

Originally Posted by Yale
How can you tell if the limit is from above or below?

Do you mean from the left or the right?

Do you mean f(x) tendind to -infinity or +infinity?

I'm having trouble understanding your question...
• Jun 11th 2009, 04:52 PM
Yale
Quote:

Originally Posted by VonNemo19
Do you mean from the left or the right?

Do you mean f(x) tendind to -infinity or +infinity?

I'm having trouble understanding your question...

Perhaps they are more commonly referred to as left or right but the text I am looking at calls them above or below. What you posted there does seem to match up however. How would you go about determining that?
• Jun 11th 2009, 04:57 PM
VonNemo19
Quote:

Originally Posted by Yale
Perhaps they are more commonly referred to as left or right but the text I am looking at calls them above or below. What you posted there does seem to match up however. How would you go about determining that?

Generally try direct substitution. If that does not work, try to rewrite the expression by simplifying, rationalizing, etc.. until direct substitution works. If the function is piece wise, pay carefull attention to which piece of the function to use with which variable. This is really hard to discuss without an example.
• Jun 11th 2009, 05:02 PM
Yale
Quote:

Originally Posted by VonNemo19
Generally try direct substitution. If that does not work, try to rewrite the expression by simplifying, rationalizing, etc.. until direct substitution works. If the function is piece wise, pay carefull attention to which piece of the function to use with which variable. This is really hard to discuss without an example.

I shall do my best to post an example then:

lim (5x^2 -6x - 1) / (2x^2 + 6)
x -> positive infinity

I know how to solve to get 5 / 2 but I am not sure if it is above or below (left or right).
• Jun 12th 2009, 03:03 AM
mr fantastic
Quote:

Originally Posted by Yale
Perhaps they are more commonly referred to as left or right but the text I am looking at calls them above or below. What you posted there does seem to match up however. How would you go about determining that?

Post the question that has prompted your question.
• Jun 12th 2009, 07:26 AM
VonNemo19
Quote:

Originally Posted by Yale
I shall do my best to post an example then:

lim (5x^2 -6x - 1) / (2x^2 + 6)
x -> positive infinity

I know how to solve to get 5 / 2 but I am not sure if it is above or below (left or right).

This is easy as long as you understand that $\displaystyle \lim_{x\to+\infty}\frac{1}{x}=0$

Intuitively this makes sense because as x becomes larger and larger $\displaystyle \frac{1}{x}$ becomes smaller and smaller.

So, whenever we can rewrite a function to look like something similar to $\displaystyle \frac{1}{x}$ then we can take the limit. Like so...

$\displaystyle \frac{5x^2-6x-1}{2x^2+6}=\frac{x^2(5-\frac{6}{x}-\frac{1}{x^2})}{x^2(2+\frac{6}{x^2})}$

$\displaystyle =\frac{5-\frac{6}{x}-\frac{1}{x^2}}{2+\frac{6}{x^2}}$

So, by making use of our limit properties, we can "distribute" the limit throughout the terms.

What happens to $\displaystyle \frac{6}{x}$ as x tends to intinity?

It goes to zero.

The same is true for the terms: $\displaystyle \frac{6}{x^2}=6\cdot\frac{1}{x}\cdot\frac{1}{x}$ as x tends to infinity we have

$\displaystyle 6\cdot0\cdot0=0$

So evaluating the limit we have

$\displaystyle \lim_{x\to+\infty}\frac{5x^2-6x-1}{2x^2+6}=\lim_{x\to+\infty}\frac{5-\frac{6}{x}-\frac{1}{x^2}}{2+\frac{6}{x^2}}=$

$\displaystyle =\frac{5-0-0}{2+0}=\frac{5}{2}$

How can you tend to positive infinty from the right?
• Jun 12th 2009, 08:11 AM
HallsofIvy
Quote:

Originally Posted by Yale
I shall do my best to post an example then:

lim (5x^2 -6x - 1) / (2x^2 + 6)
x -> positive infinity

I know how to solve to get 5 / 2 but I am not sure if it is above or below (left or right).

How can you be above infinity?

The only way you can approach infinity is from below!

IF you had something like "$\displaystyle \lim_{x\rightarrow 2} \frac{5x^2- 6x- 1}{2x^2+ 6}$", in order for that to exist you must get the same thing from both sides.

IF you have something like "$\displaystyle \lim_{x\rightarrow 2^-}\frac{5x^2- 6x- 1}{2x^2+ 6}$" that is "from below" or "from the left" because of that little "-" on the 2. IF you have something like "$\displaystyle \lim_{x\rightarrow 2^+}\frac{5x^2- 6x- 1}{2x^2+ 6}$" that is "from above" or "from the right" because of that little "+" on the 2. In this case because the limit itself exists, both are the same.