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- January 2nd 2013, 08:20 AMrozuutLimit
- January 2nd 2013, 04:04 PMchiroRe: Limit
Hey rozuut.

I'm not sure if you have to use delta-epsilon or not, but you can use the fact that 2*sin(theta)*cos(theta) = sin(2*theta) and show that if theta doesn't converge then neither does the limit.

You will have to also partition (-1)^n * n with the other term as well, but the above is my suggestion to showing that it doesn't exist (you can show that (-1)^n also doesn't exist by writing it as a trig term of cos(pi*n) = (-1)^n). - January 2nd 2013, 05:26 PMSorobanRe: Limit
Hello, rozuut!

I have some ideas to get you started . . .

Quote:

We have: .

. .

The first limit fluctuates between +1 and -1.

The second limit is the sine of a large multiple of

. . . .

The third limit is 1: .

Can you continue?