# Limit when an expoent goes to infinity

Please could someone help me and see if my solutions are correct for these two limits

Let $n \in \mathbb{N}$ and $y \in \mathbb{R}$ and $y>0$.

Case 1 $$\lim_{y \to \infty} \cfrac{y^n}{y^{n+1}}=\lim_{y \to \infty} y^{n-(n+1)}= \lim_{y \to \infty} \cfrac{1}{y}=0$$

Case 2 $$\lim_{n \to \infty} \cfrac{y^n}{y^{n+1}}$$

In this case, I consider that I cannot manipulate the exponents in the same way as in case 1. So let be

Case 2

$$L = \lim_{n \to \infty} \cfrac{y^n}{y^{n+1}}$$

$$\sqrt[n]{L} = \sqrt[n]{\lim_{n \to \infty} \cfrac{y^n}{y^{n+1}}}=\lim_{n \to \infty} \sqrt[n]{\cfrac{y^n}{y^{n+1}}}=\lim_{n \to \infty} \cfrac{y^{n/n}}{y^{(n+1)/n}}$$

$$\sqrt[n]{L} = \lim_{n \to \infty} \cfrac{y^{1}}{y^{1+1/n}}=1$$

And thus,

$$L = 1^n = 1$$

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The first step you have done is just algebra, it has nothing to do with limits. So certainly you can do it the same way in case 1 and case 2. – David Feb 25 '14 at 23:05
@David I don't think so. Algebra works quite nice when you do not involve $\infty$. In some way, I am asking $\lim_{n \to \infty}n-(n+1)$. Or also $\infty - \infty$?. – Erivelton Geraldo Nepomuceno Feb 25 '14 at 23:14
I'm only talking about the first step. This is before you take the limit and therefore it does not involve $\infty$. – David Feb 25 '14 at 23:23

No, for all positive n, case 2 is the constant 1/y. Why the doubt that you cannot "manipulate the exponents in the same way"?

For all n, $\frac{y^n}{y^{n+1}} = \frac{1}{y}$. To convince yourself of this, plug in say, 2 for n, then plug in 3, then plug in 4 etc. But you could just as easily write $\frac{y^n}{y^{n+1}} = \frac{1}{y^{(n + 1) - n}} = \frac{1}{y}$.

To see the error in the "proof" take the n-th root of any positive number C and then take the limit:

$$\lim_{n \to \infty} \sqrt[n]{C} = 1$$

And for all n, $$1^{n} = 1$$

Nevertheless, this does not prove that C = 1. (In your case 2, C is 1/y.)

Edit:

What the above argument shows is that taking the limit of the n-th root of any positive number destroys all information on what postive number you started with. Similarly, multiplying a number by zero destroys all infomation on what number you started with. Taking the n-th root and raising to the n-th power are inverses of each other, but not if you take limits in between.

Also, I agree that Case 1 is perfectly fine.

Another edit:

The particular part that is incorrect is that in fact,

$$\sqrt[n]{\lim_{n \to \infty} \cfrac{y^n}{y^{n+1}}} \neq \lim_{n \to \infty} \sqrt[n]{\cfrac{y^n}{y^{n+1}}}$$

The problem is that the $n$ outside of the limit and the $n$ inside are different. The one on the outside is a fixed number (it doesn't approach infinity). But once it is put inside the limit, it is no longer fixed. The n underneath the limit is a dummy variable. i.e.,

$$\lim_{n \to \infty} \frac{y^n}{y^{n+1}} = \lim_{j \to \infty} \frac{y^j}{y^{j+1}}.$$

It is true that

$$\sqrt[n]{\lim_{j \to \infty} \frac{y^j}{y^{j+1}}} = \lim_{j \to \infty} \sqrt[n]{\frac{y^j}{y^{j+1}}}$$

but then we see that the n and the j do not simplify.

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Let $p=n$ and $k=n+1$ $$\lim_{n \to \infty} \cfrac{p}{k} = \lim_{n \to \infty} \cfrac{n}{n+1} = \lim_{n \to \infty} \cfrac{n}{n+1}\cfrac{\frac{1}{n}}{\frac{1}{n}} = \lim_{n \to \infty} \cfrac{1}{1+\frac{1}{n}} =1$$ That is $$\lim_{n \to \infty} \cfrac{p}{k} =1$$ $p/k$ goes to $1$ if and only if $p-k$ goes to $0$. Case 2 $$\lim_{n \to \infty} \cfrac{y^p}{y^k} = \lim_{n \to \infty} y^{p-k} =\lim_{n \to \infty} y^{0} = 1$$ The result $1/y$ implies that $p/k$ does not go to $1$, which seems to me that contradicts the notion of infinity. @David ? – Erivelton Geraldo Nepomuceno Feb 26 '14 at 5:19
No, the statement p/k goes to 1 iff p-k goes to 0 is false. Let f(n) = n^2 + n and g(n) = n^2. Then f(n) - g(n) approaches infinity. Yest still f(n)/g(n) approaches 1. In your case, p-k is the constant 1. – Andrew Kelley Feb 26 '14 at 21:56
Oops. p-k is the constant -1. – Andrew Kelley Feb 27 '14 at 23:11
Dear @AndrewKelley, I am not confident with this reasoning. I found this affirmation in "Cauchi's Cours d'analyse": When variables remain independent, the values of $x/y = 0/0 = M((- \infty, + \infty))$ or $x/y = \infty / \infty =-\infty /- \infty= M((0,+ \infty))$. When I take $y^n$ I mean a function independent of $y^{(n+1)}$. And therefore, for $y<1$, it would be $M((-\infty,+ \infty))$ and for $y=1$, it would be $1$, and for $y>1$ it will be $M((0,+ \infty))$. Where $M((a,b))$ denotes any value betwen the limits of $a$ and $b$. What do you think? – Erivelton Geraldo Nepomuceno Feb 28 '14 at 13:46
I like your approach when you use destroy the information. I also think that when someone just cancel terms $n$ from two different functions, I think it would be also destroying the information about the individuality of each function. @andrew-kelley – Erivelton Geraldo Nepomuceno Feb 28 '14 at 13:53

Looks good.

For case 1, the denominator grows faster and thus the limit goes to 0.

The second case is wrong however. It can be simplified like this:

$$\lim_{n \to \infty} \frac{y^n}{y^{n+1}} = \lim_{n \to \infty} \frac{1}{y} = \frac{1}{y}$$

In other words, you can use the same method as for case 1.

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Thank you. Let us see if other guys accord with my additional argument. What do you think? – Erivelton Geraldo Nepomuceno Feb 26 '14 at 5:23
After reading the other answer I realised I was wrong about case 2. I've changed the details now. – naslundx Feb 28 '14 at 18:22
Did you add something else? @naslundx – Erivelton Geraldo Nepomuceno Mar 1 '14 at 5:46