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jverkoey
10-24-2005, 07:11 PM
<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:

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

I found two ways to go about doing this.

1) divide all by x:

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:

-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?

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:

-inf
----
+inf
Which is -inf.

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

-inf
----
-inf
Which is +inf.

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.

Thantos
10-24-2005, 07:28 PM
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:

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

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.

jverkoey
10-24-2005, 07:34 PM
Thanks Thantos.

Now I must search for that inf ascii value...

Thantos
10-24-2005, 08:32 PM
start -> run -> charmap -> advanced view -> search for "infi" -> search -> select -> copy -> alt+tab -> paste
:)

Dweia
10-24-2005, 08:43 PM
the infinity sign isn't ascii, it'd be unicode, like delta and all that.

Perspective
10-26-2005, 05:38 PM
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.

jverkoey
10-26-2005, 08:01 PM
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.

Rashakil Fol
10-26-2005, 09:45 PM
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.

sunnypalsingh
10-26-2005, 11:05 PM
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.....