Question

求函数极限如图,求极限此题答案是 正无穷大，想要过程

Answer 1

1/x^2-1/sin^2(x)=(sin^2(x)-x^2)/(x^2*sin^2(x)) (1)

expand sin(x) to the order of x^3 we have

sin(x)=x-1/6*x^3+O(x^3) (2)

so

(sin^2(x)-x^2)=[x-1/6*x^3+O(x^3)]^2-x^2 (3)

because it is devided by sin^2(x)*x^2---x^4

so we only keep the x^4 term(ignore higher orders of x)

in (3),which is

-1/3*x^4

so the final resault is -1/3

答案是错的,Maple 13软件计算了是-1/3