# 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:

$$\lim_{n\to \infty}\Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr)^n=\nexists$$

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You cannot conclude that the limit does not exist from the fact that $a_n$ and $b_n$ do not approach the same limit. Suppose you had the sequence $x_n=1/n$. Then $x_n\to 0$, but we could select $b_n=1/n$ and $a_n=-n$. – Jeff Nov 22 '11 at 21:04

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.
Or, less rigorously: As $n\to\infty$, $\frac1{n^2}\to0$ so the base $\approx\frac1k\cos n$; $|\cos n|\lt1$ so the base $\le\frac1k$ in absolute value. Taking the $n\to\infty$th power yields $0$. – msh210 Nov 22 '11 at 22:36