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,
$$ L = 1^n = 1$$
 A: 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.
A: 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. 
