Question

lim x→∞[(x^2+5x)/(x^2-5x)]^(x^k),当k取何值,极限为常数

Answer 1

超速把题改了==令y=[(x^2+5x)/(x^2-5x)]^(x^k)lny=x^k ln[(x^2+5x)/(x^2-5x)]=ln[(x^2+5x)/(x^2-5x)]/(x^(-k))若y极限是常数则lny极限也是常数用等价无穷小替换ln(1+t)~t,t->0此处所以有ln[(x^2+5x)/(x^2-5x)]=ln[1+...