請問lim(n->∞)cos(a2)cos(a4)...cos(a2^n)(a不等於0)如果想使用夾逼定理的話應怎麼做?
- 2022-09-12
lim(r->0)[1/πr2]∫∫e^(x2-y2)cos(x+y)dxdy,其中D為x2+y2≤r2 由積分中值定理,在D記憶體在點(a,b),使: ∫∫e^(x2-y2)cos(x+y)dxdy=πr2e^(a2-b2)cos(a+b) 所以:lim(r->0)[1/πr2]∫∫e^(x2-y2)cos(x+y)dxdy =lim(r->0)[1/πr2]πr2e^(a2-b2)cos(a+b) =lim(r->0)e^(a2-b2)cos(a+b) =1
0< n!/n^n = 1*2*3*。。。*n / n*n*。。。*n < 1/n
∵ lim(n→∞)1/n = 0
∴ lim(n→∞)n!/n^n = 0