Question

第9题,(3)和(5)怎么做啊.(3)我怎么算2啊,5啊.,(5)用夹逼定理两边极限不都是0了吗?答案怎么是1.

Answer 1

（3）∵lim(x->0)[x/(e^x-1)]

=lim(x->0)(1/e^x) (0/0型极限,应用罗比达法则)

=1

lim(x->0)[x/ln(1+x)]

=lim(x->0)(1+x) (0/0型极限,应用罗比达法则)

=1

lim(x->0)[(1-cosx)/x^2]

=lim(x->0)[(1/2)(sinx/x)] (0/0型极限,应用罗比达法则)

=(1/2)*1 (应用重要极限lim(t->0)(sint/t)=1)

=1/2

∴lim(x->0){(1-cosx)/[(e^x-1)ln(1+x)]}

=lim(x->0){[x/(e^x-1)]*[x/ln(1+x)]*[(1-cosx)/x^2]}

={lim(x->0)[x/(e^x-1)]}*{lim(x->0)[x/ln(1+x)]}*{lim(x->0)[(1-cosx)/x^2]}

=1*1*(1/2)

=1/2

=0.5

（5）令An=1/√(n^2+1)+1/√(n^2+2)+.+1/√(n^2+n)

∵n/√(n^2+n)≤An≤n/√(n^2+1)∞)[n/√(n^2+n)]=lim(n->∞)[1/√(1+1/n)]=1

∴1=lim(n->∞)[n/√(n^2+n)]≤lim(n->∞)An≤n/√(n^2+1)lim(n->∞)An=1

即lim(n->∞)[1/√(n^2+1)+1/√(n^2+2)+.+1/√(n^2+n)]=1.