求极限（根据有界性定理或夹逼定理）---

1个回答

[2^n+(-1)^n]/[2^(n-1)+sin(nπ/2)]
= 2[1+(-1/2)^(n-1)]/[1+sin(nπ/2)/2^(n-1)]
→2(n→+inf.)