<edit>

Bah, after re-reading my post I guess it's not a very quick question. :rolleyes: </edit>

So I just took a midterm, and one of the questions was as follows:

I found two ways to go about doing this.Code:`lim (1-x^2)`

x->+inf -------

(3+x)

1) divide all by x:

When you plug in +inf, you get:Code:`lim (1-x^2)/x`

x->+inf -------

(3+x)/x

lim (1/x-x)

x->+inf -------

(3/x+1)

Evaluating to -inf.Code:`-inf`

----

1

I checked this over at the end of the test though, and decided I didn't like this solution, as by the definition of a limit: the left side limit must equal the right side limit...and how do you evaluate the left and right side of infinity?

So, I erased that answer and proceeded to write answer #2:

So, for lim 1/x->0+, we have:Code:`lim (1-x^2)`

x->+inf -------

(3+x)

As lim lim

x->+inf 1/x->0.

Therefore,

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

1/x->0 ----------

(3+(1/x))

Which is -inf.Code:`-inf`

----

+inf

Also, for lim 1/x->0-, we have:

Which is +inf.Code:`-inf`

----

-inf

So, yah, a bunch of my friends all said -inf is right, but I really don't like the idea of putting that down as saying it is -inf implies the limit exists and approaches -inf, which isn't necessarily true, as inf isn't defined as a number....Aurgh.Code:`lim1/x->0- DNE lim1/x->0+,`

Therefore lim1/x->0 DNE.