Question

（x+(1+x^2)^1/2）^1/x的极限,x趋于正无穷

Answer 1

lim(x->正无穷)（x+(1+x^2)^1/2）^1/x

对（x+(1+x^2)^1/2）^1/x 求对数

lim(x->正无穷)ln[（x+(1+x^2)^1/2）^1/x ]

=lim(x->正无穷){ln[x+(1+x^2)^1/2]}/x (∞/∞型,适用罗必塔法则）

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

=lim(x->∞)[(1+x^2)^(1/2)+x]/{[x+(1+x^2)^1/2]*[(1+x^2)^(1/2)]}

=lim(x->∞)1/[(1+x^2)^(1/2)]}

=0

e^0=1

所以lim(x->正无穷)（x+(1+x^2)^1/2）^1/x=1