Existence of the limit of a sequence? I solved this limit problem by following this way, but I'm not exactly sure about ....
can anyone help me and tell me if it is correct?
the problem is:

Let $k>1$. If it exists, calculate the limit of the sequence $(x_n)$,
  $$x_n := \Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr)^n.$$

my solution:
From trigonometry we know that:
$$ 0< \sin x < x <\tan x$$ 
for $$0 < x < \pi/2 ;$$
then $$\sin x < x \Rightarrow \sin \frac{1}{n^2}<\frac{1}{n^2};$$
the cosine function we know to be a bounded function; then:
$$\left | \cos x \right |\le1\Rightarrow -1\le\cos x\le 1 \Rightarrow -\frac{1}{k}\le
\frac{1}{k}\cos n \le\frac{1}{k} $$ 
so we have that:
$$k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \le \Biggl(k\frac{1}{n^2}+\frac{1}{k}\Biggr) \Rightarrow\Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr)^n\le \Biggl(k\frac{1}{n^2}+\frac{1}{k}\Biggr)^n,$$
then
$$x_n \le \Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n;$$
On the other hand, we have also that: 
$$\left | \sin x \right |\le1\Rightarrow -1\le\sin x\le 1 \Rightarrow -{k}\le\ k \cdot \sin (\frac{1}{n^2}) \le{k},$$
and then in particular
$$-\frac{1}{k}\le\cos n \qquad,\qquad -{k}\le\ k \cdot \sin (\frac{1}{n^2});$$
well:
$$\Biggl(-k-\frac{1}{k}\Biggr)\le\Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr) \Rightarrow -\Biggl(k+\frac{1}{k}\Biggr)^n\le\Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr)^n,$$
end then:
$$-\Biggl(k+\frac{1}{k}\Biggr)^n\le x_n;$$
combining the two results we obtained:
$$-\Biggl(k+\frac{1}{k}\Biggr)^n\le x_n\le\Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n.$$
at this point we have found that $a_n$ $b_n$ such that:
$$a_n \le x_n \le b_n,$$
an so $x_n$, if it admits limit, must be admitted to the same limit $a_n$  and $b_n$; 
so:
$$\lim_{n\to \infty}a_n=\lim_{n\to \infty}-\Biggl(k+\frac{1}{k}\Biggr)^n=-\lim_{n\to \infty}\Biggl(\frac{k^2+1}{k}\Biggr)^n=-\infty$$
$$\lim_{n\to \infty}b_n=\lim_{n\to \infty}=\Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n=\Biggl[\frac{n^2\cdot(1+\frac{k}{n^2})}{kn^2}\Biggr]^n =\frac{\Biggl[(1+\frac{k}{n^2})^{n^2}\Biggr]}{k^n}^\frac {1}{n}=\frac {e^\frac {k}{n}}{k^n}=\frac {1}{k^n}=0$$
as seen , the two limits are different, so the sequence $ x_n $ admits no limit. In conclusion:
$$\lim_{n\to \infty}\Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr)^n=\nexists$$

The limit is zero.  You can argue as follows.  If $n$ is large enough that $1/n^2 \leq \pi/2$, then $\sin(1/n^2)\leq 1/n^2$.  Then we have that
$$|x_n| \leq \left(\frac{k}{n^2} + \frac{1}{k}\right)^n.$$
Then choose $n$ large enough so that $\frac{k}{n^2} + \frac{1}{k} \leq 1-\delta$ for some positive $\delta<1$.  This can be done as $k > 1$ so $1/k < 1$.  Then you have that 
$$|x_n| \leq (1-\delta)^n$$
for large enough $n$, and this limit is zero.
