Epsilon-Delta Proof at infinity

Question

Let $n \in \mathbb{N}$ and $y \in \mathbb{R}$ and $0<y<1$. Let also be $f(y)=y^n$ and $g(y)=y^{n+1}$.

$$ \lim_{n \to \infty} \cfrac{f(y)}{g(y)} = L $$

What is the value of $L$ using the epsilon-delta definition of limit?

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Edit

Consider $n \in \mathbb{R}$

Answer 1

Note that for any $n$,

$$ \frac{f(y)}{g(y)} = \frac{y^n}{y^{n+1}} = \frac{1}{y} $$

Now fix $\epsilon > 0$. Choose $N = 1$. For $n > N$, we have $$ \left| \frac{f_{(n)}(y)}{g_{(n)}(y)} - \frac{1}{y} \right| = \left| \frac{1}{y} - \frac{1}{y} \right| = 0 < \epsilon. $$ Q. E. D.

Answer 2

For $0<y<1$, it must be $L=1/y$

Answer 3

As $\frac{f(y)}{g(y)} = \frac{1}{y}$ and does not depend on $n$, then this limit equals exactly $\frac{1}{y}$.

Using the epsilon-delta definition of limit you can write $\forall \varepsilon > 0 \; \exists n_0 = 1: \forall n > n_0 \; \left|\frac{f(y)}{g(y)} - \frac{1}{y}\right| < \varepsilon$, because $\left|\frac{f(y)}{g(y)} - \frac{1}{y}\right| = 0$ and does not depend on $n$.