Question

求下列极限 lim(x→∞) [(x^3-x^2+x/2)*e^(1/x)-√x^6+1]

Answer 1

7/6

lim(x->∞) ((x^3-x^2+x/2)*e^(1/x)-√x^6+1)
=lim(x->∞) ((x^3-x^2+x/2)*e^(1/x)-√x^6) +1
=lim(x->∞) (x/2 e^(1/x) (2x^2-2x+1)-x^3) +1
=lim(x->∞) (x/2 (e^(1/x) (2x^2-2x+1)-2x^2)) +1
=(1/2) lim(x->∞) (x(e^(1/x) (2x^2-2x+1)-2x^2))+1
令t=1/x
得到
(1/2) lim(t->0) ((e^t (2t^(-2)-2t^(-1)+1)-2t^(-2))/t)+1
=(1/2) lim(t->0) ((e^t (t^2-2t+2)/t^3)+1
明显地，当t->0时，极限内分母趋近于0，分子也趋近于0
因此使用L'Hospital法则得
(1/2) lim(t->0) (((e^t (t^2-2t+2)+e^t (2t-1))/3t^2)+1
=(1/2) lim(t->0) (((e^t (t^2))/3t^2)+1
=(1/2) lim(t->0) (e^t/3)+1
直接代入t=0得
(1/2)*(1/3)+1
=7/6
所以
lim(x->∞) ((x^3-x^2+x/2)*e^(1/x)-√x^6+1)=7/6