lim n→∞ sin pi√(n^2+a^2) (a不等于0)

先考虑 lim n→∞ |sin [pi(n^2+a^2)^(1/2)|
lim n→∞ |sin [pi(n^2+a^2)^(1/2)|
= lim n→∞ |sin [pi(n^2+a^2)^(1/2) - npi]|
= lim n→∞ |sin {pi[(n^2+a^2)^(1/2) - n]}|
= lim n→∞ |sin {pi[(n^2+a^2) - n^2]/[(n^2+a^2)^(1/2) + n]}|
= lim n→∞ |sin {pi[a^2]/[(n^2+a^2)^(1/2) + n]}|
= |sin(0)|
= 0.
所以
lim n→∞ sin pi(n^2+a^2)^(1/2)
= 0

原式等于lim n趋于无穷 sinnπ
极限不存在

0啊

n→∞时,n^2+a^2→n^2
∴√(n^2+a^2) →n
∴pi√(n^2+a^2) →npi
则极限lim n→∞ sin pi√(n^2+a^2)
=lim n→∞ sin npi
=0