help please with number 5

Question

marcelie · Answer

attachment

solomonzelman · Answer

We know that  $-1\le \sin n \le 1$.
(And $-1\le \cos n \le 1$)

So, intuitively,
$\displaystyle \lim_{n\to\infty}\left(\frac{n}{n^2+1}\right)\sin n$

solomonzelman · Answer

is equal to 0.

marcelie · Answer

is that the squeeze theorem ? and why is the sin n outside?

solomonzelman · Answer

Oh, good that you mentioned, yes, we can verify this through squeeze trm.

solomonzelman · Answer

$\displaystyle \lim_{n\to\infty}\frac{-n}{n^2+1}\le \lim_{n\to\infty}\frac{n\sin n}{n^2+1}\le \lim_{n\to\infty}\frac{n}{n^2+1}$

solomonzelman · Answer

well, you know the limits on the left and right are equal to what ?

marcelie · Answer

hmm im sorry i got lost D:

solomonzelman · Answer

Do you see why I set

$\displaystyle \lim_{n\to\infty}\frac{-n}{n^2+1}\le \lim_{n\to\infty}\frac{n\sin n}{n^2+1}\le \lim_{n\to\infty}\frac{n}{n^2+1}$

                                 ??

solomonzelman · Answer

(because $-1\le\sin n \le 1$,   ...  right?)

marcelie · Answer

oh right

solomonzelman · Answer

Can you find the limits on the left and right ends?

solomonzelman · Answer

(You can do it intuitively, or use L'Hospital's Rule)

marcelie · Answer

hmm i was thinkin that we can multiply by the greatest common factor right ?

solomonzelman · Answer

Which common factor, where ?

marcelie · Answer

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