Question

Need help understanding the solution
Please

Answer 1

I don't understand the last 2 lines. How can they conclude it?

Answer 2

Why \(|a_n b_n| <M|a_n| \) and \(lim a_n=0\) leads to lim \(a_n b_n =0\)

Answer 3

You could think of it like squeeze theorem:
\[\lim_{n \rightarrow \infty}-Ma_{n} <\lim_{n \rightarrow \infty} a_{n}b_{n} < \lim_{n \rightarrow \infty} Ma_{n}\]
\[0 < \lim_{n \rightarrow \infty}a_{n}b_{n}< 0\]
Which we know is true since b_{n} is bounded, so a_{n}b_{n} goes to 0. I would interpret it this way since it's pretty much what they're doing.

Answer 4

If so, we assume that lim \(a_n b_n\) exists?

Answer 5

consider the sequence \(a_n b_n\) . We know nothing about it, right? we know a_n, and b_n but not a_n b_n

Answer 6

Well, I would say that fitting the conditions above implies that the limit exists as well as how it is proven. I can't prove that it has to exist given those conditions, but they're not trying to trick you here.

Answer 7

It's just another way of thinking at what they did when trying to prove it. If there were a reason to think the limit for \(a_{n}b_{n}\) didnt exist, then it would be the fault of the proof you were given in the attachment. Not only that, but these sequences are always going to infinity. So in what ways might a limit not exist? Well, because we're going to infinity, we don't have the option of fitting criteria like approaching the same point from both sides, which throws out any asymptote ideas or one-sided possibilities. The only option left for a limit to not exist would be oscillating. Well, an oscillating sequence is bounded, which is accounted for by this proof.

Answer 8

Well, of course there are limits with sharp points where they're not differentiable, but once again that idea is thrown out of consideration since we're going to infinity.

Answer 9

Got you. Thanks a lot. :)

Answer 10

No problem :)