# Computing Limits

• January 30th 2007, 01:12 PM
qbkr21
Computing Limits
When I am computing a limit where the degree of the numerator is different from the degree of the denominator or vice versa what would I multiply both by in order to solve. Also if there is a square root does it change the what I would multiply by. It's easy to solve the simple limits however I am looking for a set of rules to follow when I am trying to solve the more complex problems.

Thanks
• January 30th 2007, 01:27 PM
topsquark
Quote:

Originally Posted by qbkr21
When I am computing a limit where the degree of the numerator is different from the degree of the denominator or vice versa what would I multiply both by in order to solve. Also if there is a square root does it change the what I would multiply by. It's easy to solve the simple limits however I am looking for a set of rules to follow when I am trying to solve the more complex problems.

Thanks

Degree of the numerator is different from the denominator...
Square root changes...

There are too many possibilities here. Narrow it down or give an example of what you want!

-Dan
• January 30th 2007, 01:33 PM
qbkr21
2 Examples would be real nice

1.
Degree of the numerator is different from the denominator...

2.
Square root in the denominator
• January 30th 2007, 01:38 PM
topsquark
Quote:

Originally Posted by qbkr21
2 Examples would be real nice

1.
Degree of the numerator is different from the denominator...

2.
Square root in the denominator

I'm still not sure of what you mean. Let's try this:
$\lim_{x \to 0} \frac{x^2 + 2x + 5}{x + 1} = \frac{0^2 + 2*0 + 5}{0 + 1} = 5$
Is this what you are looking to find out how to solve?

-Dan
• January 30th 2007, 01:39 PM
qbkr21
Yes now what did you multiply the numerator and denominator by?

Not a constant but infinity
• January 30th 2007, 01:41 PM
ThePerfectHacker
Quote:

Originally Posted by qbkr21
Yes now what did you multiply the numerator and denominator by?

Not a constant but infinity

You do not multiply by anything. Since you can substitute there is no need.
• January 30th 2007, 01:45 PM
qbkr21
Rules? I need some rules to play the game...
• January 30th 2007, 02:34 PM
ThePerfectHacker
Quote:

Originally Posted by qbkr21
Rules? I need some rules to play the game...

Yes.

Math is a game of rules and no objectives.
Philosophy is a game with objectives and no rules.
• January 30th 2007, 05:28 PM
qbkr21
Could someone please tell me the rules so that I can solve a hard limit problem that goes to Infinity...
• January 31st 2007, 05:57 AM
topsquark
There are no "rules" as such, just methods that have been shown to work. And there is usually more than one way to think about them. So you have to work some problems and find what you are most comfortable with.

$\lim_{x \to \infty} \frac{x^2 - 3x + 5}{x + 10}$

Note that as x goes to infinity only the terms of largest degree will be important. So we can do this in two ways:
1) Keep only the largest degree terms in the numerator and denominator:
$\lim_{x \to \infty} \frac{x^2 - 3x + 5}{x + 10} \to \lim_{x \to \infty}\frac{x^2}{x} = \lim_{x \to \infty}x \to \infty$

2) We can reduce the order of the numerator and denominator by dividing both by the largest power term, in this case $x^2$:
$\lim_{x \to \infty}\frac{x^2 - 3x + 5}{x + 10} = \lim_{x \to \infty} \frac{1 - \frac{3}{x} + \frac{5}{x^2}}{\frac{1}{x} + \frac{10}{x^2}}$
Notice that any term $\lim_{x \to \infty}\frac{1}{x^n} \to 0$, so this becomes:
$\to \frac{1}{0} \to \infty$

These processes can be done using any degree polynomials in the numerator or denominator. Also it does NOT matter that we are taking a limit as x goes to infinity, you can use similar arguments as x approaches any number as well.

-Dan