Question

求lim(1/x^2-(arctanx)^2), x趋于正无穷

Answer 1

先取对数求极限：
lim(x→＋∞)
[ln(π/2-arctanx)]/lnx
使用洛必达法则
＝lim(x→＋∞)
[1/(π/2-arctanx)×(-1)/(1+x^2)]/(1/x)
＝－lim(x→＋∞)
1/(π/2-arctanx)×x/(1+x^2)
＝－lim(x→＋∞)
(1/x)1/(π/2-arctanx)×x^2/(1+x^2)
＝－lim(x→＋∞)
(1/x)/(π/2-arctanx)
使用洛必达法则
＝－lim(x→＋∞)
(-1/x^2)/(-1/(1+x^2))
＝－lim(x→＋∞)
(1+x^2)/(x^2)
＝－1
所以，lim(x→＋∞)
(π/2-arctanx)^(1/lnx)＝lim(x→＋∞)
e^｛[ln(π/2-arctanx)]/lnx｝＝e^(-1)