Question

极限问题微积分

极限问题微积分第二题怎么做啊

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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

Answer 2

according to Stirling's approximation,

n! ~ sqrt(2 * pi * n) * (n/e)^n

limit 10^n/n! =0, lim n^10/n! =0

so you can omit these two item in the calculation,

final answer is 10

追问

谢谢，做出来了已经