Calculus attack!

The Core RPGC Main Forum
1

Woo boy, second day in calculus class. It is serious business here.

Okay, we reviewed on series and sequences, and briefly touched on recursives. Now we’re doing limits on serieses. Help needed for the question below:

<b>t</b><i>n</i> = the <i>n</i>th root of <i>n</i>

Using the above equation of the series, I had to find <b>t</b><i>n</i> if <i>n</i> equals 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 50, 100, 500, 1000, and 10 000.

I do so, easy stuff.

Then it says:

lim <i>n</i>—> ∞

That applies to the equation of the series which was given: <b>t</b><i>n</i> = the <i>n</i>th root of <i>n</i>. I know what it means and all.

But how would I find the limit? I’m guessing it’s zero.

According to this sheet the instructor handed out; stating three rules about limit:

(I omitted the “Lim n–>∞” thing, 'cause it’s too long to put in, but assume it’s infront of each of the equations below)

R.1: 1/<i>n</i>^<u>r</u> , if <u>r</u> > 0 , then it equals 0.

R.2: <u>r</u>^<i>n</i> , if |r| (what does this mean?) < 1 , then it also requals 0.

R.3: |<u>r</u>^<i>n</i>| , if |r| > 1 , then it equals ∞.

But, it doesn’t state anything about roots.

Thanks for your help.

2

|r| means the absolute value of r, or its distance from 0. So no matter what r is, it’s given as a positive.
If r = 3, |r| = 3.
If r = -4, |r| = 4, because it’s 4 away from 0.

Also, r^n does indeed state something about roots.
r^0.5 is the same as saying the square root of r. So if |n| is less than 1 (and non-zero), it’s taking some sort of root.

3

So r^(1/n) is nth root?

4

Yeah, r^1/3 is cube root, r^1/4 is fourth root, and so on.

5

So that means the limitation for the equation is 0?

6

If it’s the nth root of n, and n is infinity, that’d give Inf^(1/Inf) (yeah I don’t know the symbol for infinity) and I think that gets into L’Hopital’s Rule, or something like that for weirdo multiple infinity problems.

7

N isn’t infinity. At least I don’t think so. My instructor said that “lim n —> infinity” means it approaches infinity, but never reaches it; or some bizarre math terms.

8

Keep in mind though, there are two numbers approaching infinity in the problem.
As the base approaches infinity, the number becomes infinitely large, but as the denominator of the exponent approaches infinity, the number becomes infinitely small. I honestly don’t remember the formula for dealing with that.

9

Well, since it follows the r^n = 0, if |r|<1, I’ll just put 0. I don’t think the instructor is expecting something from a grade 11 student in a grade 12 course that would be taught later on anyways. Well, even anybody, not just me.

But, also, you told me what |r| means, but what about |r^n|?

edit; if it become infinitely bigger & smaller, isn’t it the same as zero anyways? +infinity and -infinity together makes 0, doesn’t it?

10

I’m very inclined to say the answer is 1 though, because as the exponent grows closer to zero, the number grows closer to 1.

Since r^0 = 1, r^(very close to 0) would be very close to 1. And you can’t get much closer to 0 than (1/infinitely large number).

Someone needs to come in who actually knows proofs.

11

The function you are describing, the nth root of n is n^(1/n). Adding together infinity and negative infinity is an indeterminate form since the result depends on how fast the actual terms in the original function where increasing. n^(1/n) is nowhere near r^n, so I have no idea why you are trying to apply that. And they might just expect you to find the limit numerically, by looking at the elements in the series you found (which are clearly approaching a certain non-zero number…)

12

Original Equation: y = x^(1/x)

Using a trick:

ln(y) = ln(x^(1/x)) = 1/x*ln(x) = ln(x) / x

Applying L’hopital’s Rule:

lim(x->∞)ln(y) = (1/x) / 1 = 1/x
lim(x->∞)y = e^(1/x)

Since x->∞, 1/x approaches zero, and therefore y approaches e^0, which equals 1. Much easier to just look at a graph, but here is the proof.