What's $\lim\limits_{n\to\infty} \frac{x^n}{y^n + 1}$ For $x,y \in \mathbb{R}$, x > 1, y >0 what's the limit of $\frac{x^n}{y^n + 1}$?
If $x = y \implies x^n = y^n: $
Let $\varepsilon > 0, y^N > \frac{1}{\varepsilon}, N \in \mathbb{N}$
$|\frac{x^n}{y^n + 1} - 1| = |\frac{x^n}{y^n + 1} - \frac{y^n+1}{y^n+1}| = |\frac{x^n - y^n + 1}{y^n + 1}| = |\frac{1}{y^n+1}| = \frac{1}{y^n+1} < \frac{1}{y^n} < \frac{1}{y^N} < \varepsilon\ \forall n>N \implies \lim\limits_{n\to\infty}\frac{x^n}{y^n + 1} = 1$
But what if $x<y$?
I know that the limit is 0, so I need to show that $|\frac{x^n}{y^n + 1} - 0| < \dots < \varepsilon$. I know $x < y \implies x^n < y^n \implies \frac{x^n}{y^n} < 1$ which is probably useful.
And how to prove that if $x > y$ the limit is $\infty$. I probably need to show that $\forall a \in R \exists\varepsilon > 0: |a_n-a|>\varepsilon$ with using $\frac{x^n}{y^n}>1$
Am I on the right track here?
 A: Assume $y<1$. Then the nominator goes to infinity while the denominator goes to $1$. 
Now assume $y>1$. Write the quotient as $\frac{x^n}{y^n} \frac{1}{1+\frac{1}{y^n}}$. Since $y>1$ whe have that $\frac{1}{1+\frac{1}{y^n}}$ goes to $1$. So we only have to worry about $\frac{x^n}{y^n}$. If $x>y$ we see that the limit diverges and if $y>x$ the limit is $0$. If $x=y$ then the limit is $1$.
A: Divide numerator and denominator by $x^n$:
$$
\lim_{n\to\infty}\frac{x^n}{y^n+1}=
\lim_{n\to\infty}\frac{1}{\dfrac{y^n}{x^n}+\dfrac{1}{x^n}}
$$
Since $x>1$, we know that $\lim\limits_{n\to\infty}\dfrac{1}{x^n}=0$.
Thus we have three cases:


*

*$0<y<x$: the limit is $\infty$

*$y=x$: the limit is $1$

*$y>x$: the limit is $0$


Note that for proving cases 1 and 2 it's sufficient to establish that, for $0<c<1$, $\lim\limits_{n\to\infty}c^n=0$. You can use the Bernoulli inequality for this: write
$$
c^{-1}=1+t
$$
with $t>0$; then $c^{-n}=(1+t)^n\ge1+nt$, so
$$
c^n\le\frac{1}{1+nt}
$$
and the result is proved by the squeeze theorem.
For proving case 3 you need to prove that, for $d>1$, $\lim\limits_{n\to\infty}d^n=\infty$. Again, set $d=1+t$, with $t>0$; then $d^n=(1+t)^n\ge1+nt$.
A: Not positive if this is correct, because L'hospital's rule can sometimes be fickle, but it seems to come out to:
$$\lim_{n \to \infty}\frac{x^n}{y^n + 1} = \lim_{n \to \infty}\left(\frac{\ln(x)}{\ln(y)}\right)^n$$
And thus converges to $0$ when $x < y$, diverges otherwise.
