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Sonia Gay 2022-06-22 Answered
How to find the limit of lim n ( n sin n π 2 cos 1 n )
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Answers (1)

Punktatsp
Answered 2022-06-23 Author has 22 answers
The basic strategy for showing that limits of this kind do not exist is to approach infinity along a different path; usually we consider what happens as n approaches from the evens and the odds. I've provided an example of how to do this with your problem as provided below.
First note that as n , 1 / n 0 so cos ( 1 / n ) 1; moreover, since cos ( 1 / n ) = cos ( 1 ) > cos ( π / 2 ) = 0 we have that cos ( 1 / n ) > 0 for every n 1. Now note that
sin ( n π / 2 ) = { ± 1 n 1 , 3 mod 4 0 n 0 , 2 mod 4
so if we take n with n=2m then
2 m sin ( 2 m π 2 ) cos ( 1 2 m ) = 2 m cos ( 1 2 m ) sin ( m π ) = 0
while if we take n with n odd, then
| n sin ( n π 2 ) cos ( 1 n ) | = | ( 2 k + 1 ) sin ( ( 2 k + 1 ) π 2 ) cos ( 1 2 k + 1 ) | = | ( 2 k + 1 ) cos ( 1 2 k + 1 ) |
which evidently approaches as n . Since we have approached along two different paths (namely n even and n odd) that produce different values, we conclude that the limit does not exist
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