Year 12s: what will be the first way you research your future options? >>

LImits tending to infinity/squeeze theorem

l001990lou

could some please help me
i know that to do this question i need to use the squeeze result and i understand that theorem when it doesnt involve limiting to infinity or when it doesnt iinvolve trig

te question is :

Show that lim n-->infinity [sin(n)] / n =0

i would use the coded thing but not sure how to and havent got the time to do so so sorry if you dont undertand it properly :P

Reply 1

v-zero

Squueze between -1/n and 1/n, since sin(n) will never be less than -1 or greater than 1.

Reply 2

l001990lou

yeah thats the problem , i got that far

so far it says...

lim (-1/n) < lim [sin(n)]/n < lim (1/n)

what'd be the next step?

Reply 3

l001990lou

v-zero

Squueze between -1/n and 1/n, since sin(n) will never be less than -1 or greater than 1.

yeah thats the problem , i got that far

so far it says...

lim (-1/n) < lim [sin(n)]/n < lim (1/n)

what'd be the next step?
Edit/Delete Message

Reply 4

v-zero

l001990lou

yeah thats the problem , i got that far

so far it says...

lim (-1/n) < lim [sin(n)]/n < lim (1/n)

what'd be the next step?

Now state that the limit of 1/n and -1/n at infinity are both zero, and hence by the squeezing theorem so is that of sin(n)/n ...

Wait, your statement is wrong. You've stated something about limits, which isn't what you're trying to do. You're meant to be stating that for all n you have -1/n is lessthan or equal to sin(n)/n which is less than or equal to 1/n. The squeeze theorem is a statement about each point of a sequence, not just the sequence at infinty or another point.