Question

函数极限问题？

一道函数极限问题，求解答

Answer 1

lim(x->0) [(3+tanx)^x -(3+sinx)^x]/ { x^2.【√(1+xtan2x) -1】 }
有理化分母

=lim(x->0) [(3+tanx)^x -(3+sinx)^x].[√(1+xtan2x) +1]/ {x^2.[(1+xtan2x) -1] }
=lim(x->0) [(3+tanx)^x -(3+sinx)^x].[√(1+xtan2x) +1]/ (x^3.tan2x)
=2lim(x->0) [(3+tanx)^x -(3+sinx)^x]./ (x^3.tan2x)
=lim(x->0) [(3+tanx)^x -(3+sinx)^x]/ x^4
=lim(x->0) (3^x).[ (1+tanx/3 )^x -(1+sinx/3)^x ]/ x^4
=lim(x->0) [ (1+tanx/3 )^x -(1+sinx/3)^x ]/ x^4
=lim(x->0){【(1+tanx/3 )^(3/tanx)】^(xtanx/3)-【(1+sinx/3)^(3/sinx)】^(xsinx/3)}/ x^4
=lim(x->0)[ e^(xtanx/3)-e^(xsinx/3) ]/ x^4
=lim(x->0){e^【(x/3)(x+(1/3)x^3+o(x^3))】-e^【(x/3)(x-(1/6)x^3+o(x^3))】}/ x^4
=lim(x->0) e^(x^2/3).{e^[(1/9)x^4+o(x^4)]-e^[-(1/18)x^4+o(x^4)] }/ x^4
=lim(x->0) { e^[(1/9)x^4+o(x^4)]-e^[-(1/18)x^4+o(x^4)] }/ x^4
=lim(x->0) [ ( 1+ (1/9)x^4 +o(x^4))- ( 1- (1/18)x^4 +o(x^4))]/ x^4
=lim(x->0) (1/6)x^4/ x^4
=1/6