Question

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Can you teach me how to approach these kind of problems? I know you have to take the limit and see if it converges but I'm not sure what n = ? means..

Answer 1

first one is bounded because sine is. that is, \(-1\leq \sin(x)\leq 1\)

Answer 2

second one, this is identical to asking for \(\lim_{x\to \infty}\frac{x}{\ln(x)}\) if the answer is not obvious, you can use l'hoptal's rule. both numerator and denominator grow without bound, that is they both go to infinity.
but x goes much faster than \(\ln(x)\) so there is no limit, i.e. the sequence is unbounded
are you familiar with l'hopital?

Answer 3

yeah i know l'hopital..but what does n = 2 mean for the second one?

Answer 4

third one: as \(n\to \infty\), then \(\frac{1}{n}\to 0\) and so you get \(e^0=1\) as a limit, so it is bounded

Answer 5

oh they start at n = 2 because you cannot start at n = 1

Answer 6

if you replace n by 1 in the second one you would get \(\frac{1}{\ln(1)}=\frac{1}{0}\) which is undefined

Answer 7

that is why they are starting at n = 2 and not n = 1

Answer 8

oh i see. Is the last one 0 then?

Answer 9

the limit is zero, yes, so the sequence is bounded

Answer 10

okay thanks!

Answer 11

yw