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$$\lim _{n \to \infty} \frac{n^{\beta}}{n^{\alpha + \sin(n)}} = \lim _{n \to \infty} n^{\beta - \alpha - \sin(n)} $$

if $\beta - \alpha < -2 $

the limit is clearly $0$

if on the other hand $\beta - \alpha > 2$

limit is $+\infty$

for the case $\beta = \alpha = 0$ I think the limit doesn't exist

but I can't see how the limit could possibly be $1$ as $\beta - \alpha -\sin(n)$ acts randomly for big $n$'s so it's no way steadily near $0$

amWhy
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the_firehawk
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2 Answers2

1

Notice that $\{\sin(n)\,|\,n\in\mathbb N\}$ is dense in $[-1,1]$. This means the set of limit points of the sequence $\big(\sin(n)\big)_{n\in\mathbb N}$ is $[-1,1]$, ie, for any $x\in[-1,1]$ there is a subsequence $n_k$ with $\sin(n_k) \to x$ as $k\to\infty$.

For details, see this question. Basically, it is a consequence of the fact that any additive group in the reals is either discrete of the form $a\mathbb Z$, or else dense (and this in turn follows basically from division with remainder).

It follows that $\{\beta - \alpha - \sin(n)\,|\,n\in\mathbb N\}$ has limit set $L=[\beta-\alpha-1,\beta-\alpha+1]$. We hence have the cases:

  • If $\beta-\alpha-1>0$, then $n^{\beta-\alpha-\sin(n)}>n^{\beta-\alpha-1}$. Since the latter goes to $+\infty$ as $n\to+\infty$, so does the former.

  • If $\beta - \alpha +1 < 0$, then $0<n^{\beta-\alpha-\sin(n)}<n^{\beta-\alpha+1}$. Since the latter goes to $0$ as $n\to+\infty$, by the squeeze theorem, so does the former.

  • If $\beta-\alpha-1 = 0$, then $L$ contains positive elements as well as $0$. In this case, we can find subsequences that diverge to $+\infty$ or go to $1$ as $n\to+\infty$. In particular, the limit does not exist.

  • If $\beta-\alpha+1 = 0$, then $L$ contains negative elements as well as $0$. In this case, we can find subsequences that go to $0$ or go to $1$ as $n\to+\infty$. In particular, the limit does not exist.

  • Finally, if $-1<\beta-\alpha< 1$, then $L$ contains both positive and negative elements, as well as $0$. In this case, we can find subsequences that diverge to $+\infty$, go to $0$ or go to $1$ as $n\to+\infty$. In particular, the limit does not exist.

To specifically answer your question, the limit is never $1$, but sometimes it is guaranteed to be $0$ or $+\infty$.

Fimpellizzeri
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0

To have limit equal to $1$ we need as necessary condition that

$${\beta - \alpha - \sin(n)}\to 0$$

which is never true since that limit doesn't exist.

user
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