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Need help finding Limit of sequence

(n411n+n99n72n+1) (\frac{n^411^n + n^99^n}{7^{2n}+1})
I know it tends to 0 but I can't think how to explain it other than saying "72n 7^{2n} tends to infinity faster than 11n 11^n and 9n 9^n ."



I got it now, just used the ratio rule and sum rule. sorry. You can delete this thread
(edited 10 years ago)
Reply 1
Avatar for james22
james22
16
You can use the 'squeeze' theorem. If you can show that this sequence is greater than 0 (obvious) and less than another squence which tends to 0, then it must tend to 0.
Reply 2
Avatar for hello calum
hello calum
OP
10
Original post by james22
You can use the 'squeeze' theorem. If you can show that this sequence is greater than 0 (obvious) and less than another squence which tends to 0, then it must tend to 0.


Oh okay, thanks for the reply. I got an answer using the ratio lemma and sum rule but your way sounds quicker so I'll try it
Reply 3
Avatar for danny111
danny111
3
Original post by james22
You can use the 'squeeze' theorem. If you can show that this sequence is greater than 0 (obvious) and less than another squence which tends to 0, then it must tend to 0.


Is that the sandwich theorem?
Reply 4
Avatar for james22
james22
16
Original post by danny111
Is that the sandwich theorem?


Sounds like it, the word squeeze was used when I first learnt it.

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