取倒数,得
原式=1/[lim[(x^2)+1)/x^2]^x x趋近无穷大]
=1/[lim[(1+1/x^2]^x x趋近无穷大]
=1/e^lim (x->∞)x/x^2
=1/e^0
=1
lim((x^2)/(x^2)+1)^x x趋近于无穷大
取倒数,得
原式=1/[lim[(x^2)+1)/x^2]^x x趋近无穷大]
=1/[lim[(1+1/x^2]^x x趋近无穷大]
=1/e^lim (x->∞)x/x^2
=1/e^0
=1