When does this limit equal $1$? $\;\;\lim _{n \to \infty} n^{\beta - \alpha - \sin(n)}$ $$\lim _{n \to \infty} \frac{n^{\beta}}{n^{\alpha + \sin(n)}} = \lim _{n \to \infty} n^{\beta - \alpha - \sin(n)} $$
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 $0$
 A: 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 $+\infty$.
A: 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.
