Thread: Very quick math question

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

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

Code:

lim      (1-x^2)
x->+inf  -------
          (3+x)

I found two ways to go about doing this.

1) divide all by x:

Code:

lim      (1-x^2)/x
x->+inf  -------
         (3+x)/x

lim      (1/x-x)
x->+inf  -------
         (3/x+1)

When you plug in +inf, you get:

Code:

-inf
----
  1

Evaluating to -inf.

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:

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))

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

Code:

-inf
----
+inf

Which is -inf.

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

Code:

-inf
----
-inf

Which is +inf.

Code:

lim1/x->0- DNE lim1/x->0+,
Therefore lim1/x->0 DNE.

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.