# Very quick math question

• 10-24-2005
jverkoey
Very quick math question
<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:

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?

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.
• 10-24-2005
Thantos
Code:

```lim      (1-x^2) x->+∞  -------           (3+x)```
Since you are talking about x as it gets really large (becomes unbounded) you can ignore the parts that don't contain x so you basically get:
Code:

```lim      -x^2 x->+∞  -------           x```
which can then reduced to
Code:

```lim      -x x->+∞```
then you plug in ∞ and you get -∞. At this point all you are saying is that as x becomes unbounded in the positive direction the results diverge in the negative direction. Since it doesn't converge to a number the limit does not exist.
• 10-24-2005
jverkoey
Thanks Thantos.

Now I must search for that inf ascii value...
• 10-24-2005
Thantos
start -> run -> charmap -> advanced view -> search for "infi" -> search -> select -> copy -> alt+tab -> paste
:)
• 10-24-2005
Dweia
the infinity sign isn't ascii, it'd be unicode, like delta and all that.
• 10-26-2005
Perspective
Thantor is correct. But also,

In your reasoning above jver, you said a couple times that inf/inf = something, somthing. This isn't true, inf/inf is undefined. You need to apply L'hopitals's rule to evauluate it.
• 10-26-2005
jverkoey
Quote:

Originally Posted by Perspective
Thantor is correct. But also,

In your reasoning above jver, you said a couple times that inf/inf = something, somthing. This isn't true, inf/inf is undefined. You need to apply L'hopitals's rule to evauluate it.

Ahh, yes, stupid mistake on my part. So I guess the first method is correct and just using the fact that it approaches infinity implies that it doesn't exist.
• 10-26-2005
Rashakil Fol
Quote:

Originally Posted by jverkoey
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?

You don't. lim x->infinity is different notation and has a mildly different meaning than lim x->c, where c is some real number.
• 10-26-2005
sunnypalsingh
Quote:

Originally Posted by jverkoey
Ahh, yes, stupid mistake on my part. So I guess the first method is correct and just using the fact that it approaches infinity implies that it doesn't exist.

I wouldn't say infinity doesn't exist...i would say it's unreachable.....
Actually infinity is a comparitive term.....