1
$\begingroup$

I must prove that the limit

$$\lim_{n\to\infty} \left(\sin \frac{\pi n}{2} \cdot \cos \left(\sin \frac{1}{n} \right ) \right)$$

doesn't exist and also find all of its partial limits. Apparantely this question has to do with continuity of functions. Usually, when I need to prove certain (complicated) limit of a sequence, for example, $\lim_{n\to\infty} \sqrt{\frac{n+1}{2n-4}}$ I first prove that $\frac{n+1}{2n-4} \to \frac{1}{2}$ and then use the fact that $\sqrt{x}$ is continuous at $\frac{1}{2}$ together with Heine's definition to show that the limit equals to $\sqrt{\frac{1}{2}}$. But what can I do here? Any suggestions?

$\endgroup$

3 Answers 3

0
$\begingroup$

Hint:

The right hand side has a clear limit (what is it?), it remains therefore to see what values $\sin \pi n/2$ can take on (Remember that $n$ is an integer).

Does this help you see why partial limits exist?

$\endgroup$
2
  • $\begingroup$ Yes, I already noticed that $\cos \sin (1/n) \to 1$. For odd values of $n$ we have $\sin \pi n /2 = \pm 1$ and for the evens $0$, correct? And because all of these different subsequences cover the original sequence, $1,0,-1$ are all the partial limits, correct? $\endgroup$
    – Lgn
    Commented May 23, 2015 at 18:43
  • $\begingroup$ @Lgn - yes, that is correct. $\endgroup$ Commented May 23, 2015 at 19:26
0
$\begingroup$

We know that the limit of a product is the product of limits, so we can separately consider $\lim_{n\to\infty} \frac{\sin{\pi n}}{2}$ and $\lim_{n\to\infty} \cos[\sin(1/n)]$.

Since $n$ is an integer, the first limit oscillates between $-1$ and $1$. The second limit approaches 1 (as $n\to\infty$, $1/n\to 0$ so $\sin(1/n)\to 0$ and thus $\cos[\sin(1/n)]\to 1$. Therefore the product must oscillate between $-1$ and $1$, so the limit does not exist.

$\endgroup$
0
$\begingroup$

A sequence converges if and only if all of its subsequences converge.

Consider the subsequence, e defined by $a_{f(n)}$, where $f(n)=2n+1$. This subsequence coincides with the sequence

$$a_n=(-1)^n\cos \left (sin \left (\frac{1}{2n+1} \right ) \right )$$

Which does not have limit.

Notice that some sequences actually do have a limit: for instance, consider the subsequence $f(n)=2n$. For even $n$, the sequence is constantly zero because $\forall n \in \mathbb{N} $ $$\sin \left (\frac{2n\pi}{2} \right ) =\sin{\pi n}=0 $$ This subsequence clearly converges to $0$.

$\endgroup$
1
  • $\begingroup$ Yes, I understand. I'm just a bit puzzled why this question appears in my textbook in a section about continuous functions. $\endgroup$
    – Lgn
    Commented May 23, 2015 at 18:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .