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:

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

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I found two ways to go about doing this.

1) divide all by x:

Code:

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

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

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When you plug in +inf, you get:

Code:

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-inf
----
  1

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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:

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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))

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

Code:

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-inf
----
+inf

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Which is -inf.

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

Code:

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-inf
----
-inf

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Which is +inf.

Code:

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

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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:

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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:

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lim      -x^2
x->+∞  -------
          x

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which can then reduced to

Code:

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lim      -x
x->+∞

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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.

Thanks Thantos.

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

start -> run -> charmap -> advanced view -> search for "infi" -> search -> select -> copy -> alt+tab -> paste
:)

the infinity sign isn't ascii, it'd be unicode, like delta and all that.

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.

Quote:

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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.

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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.

Quote:

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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?

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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.

Quote:

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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.

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