- #36
Rogerio
- 404
- 1
dextercioby said:NO,nonononono,the limit does not exist.I didn't change my mind.
[tex] \lim_{n\rightarrow +\infty} \sin(2\pi\sqrt{n^{2}+1}) [/tex]
does not exist...
Sorry, but the limit who doesn't exist is
[tex] \lim_{n\rightarrow +\infty} \sqrt{n^2+1} [/tex]
Take a look:
[tex] \lim_{n\rightarrow +\infty} \sin(2\pi \sqrt{n^{2}+1}) = [/tex]
[tex] \lim_{n\rightarrow +\infty} \sin(2\pi (int(\sqrt{n^{2}+1}) + frac(\sqrt{n^{2}+1})) ) = [/tex]
[tex] \lim_{n\rightarrow +\infty} \sin(2\pi frac(\sqrt{n^{2}+1}) ) = 0[/tex]