一道数学题lim(n→∞)nsin(2π√(n^2+1))
3个回答

第1题应该是“lim(n→∞)2^nsin(x/2^n)=?”.

1.原式=lim(n→∞)[sin(x/2^n)/(1/2^n)]

=x*lim(n→∞)[sin(x/2^n)/(x/2^n)]

=x*1 (应用重要极限lim(x->0)(sinx/x)=1)

=x

2.原式=lim(n->∞)[(1+2/(2n+1))^(n+1)]

=lim(n->∞)[(1+2/(2n+1))^(((2n+1)/2)*((2n+2)/(2n+1)))]

=e^{lim(n->∞)[(2+2/n)/(2+1/n)]} (应用重要极限lim(x->0)[(1+x)^(1/x)]=e)

=e^1

=e.