Question

函数极限问题？

一道函数极限问题

Answer 1

(6)
lim(x->0) [f(x)-1]/x^2 = 1 (0/0)
=> f(0) =1
lim(x->0) [f(x)-1]/x^2 = 1 (0/0 分子分母分别求导)
lim(x->0) f'(x)/(2x) = 1 (0/0)
=> f'(0) =0
lim(x->0) f'(x)/(2x) = 1 (0/0 分子分母分别求导)
lim(x->0) f''(x)/2 = 1
f''(0) =2
ie
f(0) =1 and f'(0) =0 and f''(0) =2
lim(x->0) [f(arctanx)-1]/[f(arcsinx)-ln(e+(sinx)^2)]

=lim(x->0) [f(x)-1]/[f(x)-ln(e+x^2)] (0/0 分子分母分别求导)
=lim(x->0) f'(x)/[f'(x)-2x/(e+x^2)] (0/0 分子分母分别求导)
=lim(x->0) f''(x)/{f''(x)-2[(e+x^2)-2x^2)/(e+x^2)^2}
=f''(0)/[f''(0)-2e/e^2 ]
=2/(2- 2/e)
=e/(e-1)