Check out some similar questions!

DiscusDiva · May 30, 2007, 05:53 PM

I understand how to solve for a limit, but I have no clue what to do when the limit is positive or negative infinity. It seems so random to me. HELP!!
Examples
lim 3/x^2
x->00

lim 2x^2 - 6/(x-1)^2
x->00

lim x^2/ (1-x^2)
x->00

lim x^5/ (x^3-2x+6)
x->00

ebaines · May 31, 2007, 09:01 AM

Normally in limits you think about the value of the function as x approaches the limit A from both above and below its value (i.e. from the left and the right), and consequently a value $\delta$ such that $0<|x-A|<\delta$ . But in the case of infinity you can only approach from below, and this concept of the $\delta$ doesn't apply. Instead you need to find a value for S such that for any x>S the value of 3/(x^2) is less than $\epsilon$ from the limit. In the first example you gave, if X approaches infinity, what is the value of 3/(x^2)? You can see that the bigger x gets, the value of 3/(x^2) gets closer and closer to 0. So that's the limit.

More formally, from the definition of a limit given here: Limit of a function - Wikipedia, the free encyclopedia
the limit (L) in the first example is 0, and so for any $\epsilon>0$ you can find a value S such that $| \frac3 {x^2} - 0| < \epsilon$ for $x>S$ .

galactus · May 31, 2007, 11:15 AM

When finding the limits of rational expressions, you can eliminate all but the highest powers of x and then take the limit.

For instance, #2:

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

Eliminate all but the highest powers and you have:

$\lim_{x\to\infty}\frac{2x^{2}}{x^{2}}=2$

See? Now you try the other two.

asterisk_man · May 31, 2007, 11:35 AM

Don't forget that you can use l'Hopital's rule

galactus · May 31, 2007, 12:31 PM

Yea, you could, but L'Hopital is a little overkill with rational expressions. Resort to it when other methods aren't practical. Rational expressions, like those posted, are some of the easiest limits to find. Just use the highest powers.

For instance, the 3rd one is $\lim_{x\to\infty}\frac{x^{2}}{-x^{2}+1}$

Which is nothing more than $\frac{x^{2}}{-x^{2}}=-1$

With L'Hopital we diiferentiate top and bottom and get:

$\frac{2x}{-2x}=-1$ . Same thing.