lim(x趋于无穷)x(1+1/x)^x-ax-b=0求常数a、b

Question

lim(x趋于无穷)x(1+1/x)^x-ax-b=0 求常数a、b

Answer 1

lim(x趋于无穷)[x(1+1/x)^x-ax-b]/x=0=lim(x趋于无穷)(1+1/x)^x-a
a=lim(x趋于无穷)(1+1/x)^x=e
lim(x趋于无穷)x(1+1/x)^x-ex-b=0
b=lim(x趋于无穷)x(1+1/x)^x-ex=lim(t->0)[(1+t)^(1/t)-e]/t 分子分母分别求导
=lim(t->0)[ 1/(1+t)(t)-ln(1+t)/ t^2] *[1+t)^(1/t)]
ln(1+t)=t-x^2/2+x^3/3+...
取前2项代进去得 lim(t->0)[ 1/(1+t)(t)-ln(1+t)/ t^2]=-1/2
因此b=-1/2*lim(t->0)[1+t)^(1/t)]=-e/2
a=e b=-e/2