Limit when an expoent goes to infinity

Question

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,

$$ L = 1^n = 1$$

Answer 1

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.

Answer 2

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.