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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?

Clayton Louden
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    Definitely on the right track. There are some details to fill in, and a typo in the "But what if $x\lt y$" section, you want to say $\frac{x^n}{y^n}\lt 1$. Now for example to make this small, you want $(y/x)^n\gt 1/\epsilon$. If you are allowed logs, you can say this will be true if $n\gt \ln(1/\epsilon)/\ln(y/x)$. If this exercise is before logs have been officially introduced, another method will be needed. – André Nicolas Dec 08 '15 at 00:00
  • Well I coult use logs, but then I'd have to prove that first. I already have a bunch of lemmata proving simple things like $0 < a < b \implies a^{-1}>b^{-1}$, which I need a couple of times in the whole problem. This is just a very small fraction and I probably don't want to hassle with proving the l'hopital rule first. But thank you for your input. Fixed the typo :) – Clayton Louden Dec 08 '15 at 00:06
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    Suppose for example $y/x\gt 1$. Let $y/x=1+t$. By the Bernoulli Inequality $(1+t)^n\ge 1+nt$. That will let you produce a simple $N$ beyond which $(1+t)^n\gt 1/\epsilon$. Or you can use the binomial theorem to prove the same inequality. That will bypass logs. – André Nicolas Dec 08 '15 at 00:11
  • Thank you! That got me to: If $x < y \implies \dfrac{y}{x} > 1 \implies \dfrac{y}{x} := 1+t:\ \left( \dfrac{y}{x} \right)^n \geq 1+nt \implies \left( \dfrac{y}{x} \right)^{-n} \leq \dfrac{1}{1+nt} \implies \left( \dfrac{x}{y} \right)^n \leq \dfrac{1}{1+nt}$

    And: Let $\varepsilon > 0, N > \dfrac{1}{\varepsilon}\in \mathbb{N}$

    $\left| \frac{x^n}{y^n+1} \right| = \frac{x^n}{y^n+1} \leq \frac{x^n}{y^n} \stackrel{(\star)}{\leq} \frac{1}{1+nt} \leq \frac{1}{n} < \frac{1}{N} < \varepsilon\ \forall n > N$

    Leaves the prove for divergence in case $x>y$

     – Clayton Louden Dec 08 '15 at 00:39
  • You are welcome. For the case $y\lt x$ (limit is $\infty$) one has to be a little careful, because $\frac{x^n}{y^n+1}$ is smaller than $x^n/y^n$. (But not enough smaller to spoil things.) I expect you can handle it. If it is really needed I can write a partial answer. – André Nicolas Dec 08 '15 at 00:56
  • Ok, little hint please. Can't figure it out. – Clayton Louden Dec 08 '15 at 01:13
  • If $y\le 1$, then $\frac{x^n}{y^n+1}\ge \frac{x^n}{2}$. If $y\gt 1$, then $\frac{x^n}{y^n+1}\gt \frac{x^n}{2y^n}$. This, along with earlier Bernoulli Inequality stuff, will let you produce $N$ given $\epsilon$. – André Nicolas Dec 08 '15 at 01:20
  • That's odd. I can see that $\frac{x^n}{y^n+1} \geq 1+tn$ but to show divergence, I'd have to proof that $\forall a\in \mathbb{R}\exists \varepsilon > 0: |a_n - a| > \varepsilon$. Hmm... – Clayton Louden Dec 08 '15 at 01:46
  • Sorry about the $\epsilon$, I was thinking in terms of $1/\epsilon$. But instead let us say that with the information, you can now produce, for any given $B$, an $N$ such that our $n$-th term is $\gt B$ whenever $n\gt N$. – André Nicolas Dec 08 '15 at 01:53
  • I used that $\frac{x^n}{y^n+1} \geq 1+tn$ implies that $\frac{x^n}{y^n+1}$ bounded (archimedic axiom), that means, $\frac{x^n}{y^n+1}$ doesn't converge. Thanks a bunch! – Clayton Louden Dec 08 '15 at 01:59
  • The writeup, if you want all details right, will have to be careful, with all inequalities vetted. – André Nicolas Dec 08 '15 at 02:07

3 Answers3

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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 $1$.

math635
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  • I do know the limits. I'm trying to prove them or prove divergence, i.e. show that for every $\varepsilon > 0$ $|\frac{x^n}{1+y^n}| < \varepsilon$ (if $x < y)$ Thanks for the comment though. – Clayton Louden Dec 08 '15 at 00:00
  • so it is not clear why for example $\frac{x^n}{y^n}$ goes to zero if $y>x$? – math635 Dec 08 '15 at 00:05
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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:

  1. $0<y<x$: the limit is $\infty$
  2. $y=x$: the limit is $1$
  3. $y>x$: 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 $d^n=(1+t)^n\ge1+nt$.

egreg
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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.

Kaynex
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