Question

考研数学求极限这道题怎么做？

Answer 1

x->0

分子

√(1+x^2) = 1+ (1/2)x^2 +o(x^2)

x+√(1+x^2) =1+x+ (1/2)x^2 +o(x^2)

ln[x+√(1+x^2)]

=ln[1+x+ (1/2)x^2 +o(x^2)]

=[x+ (1/2)x^2] -(1/2)[x+ (1/2)x^2]^2 +o(x^2)

=[x+ (1/2)x^2] -(1/2)[x^2+o(x^2)] +o(x^2)

= x +o(x^2)

ln(1+x) = x -(1/2)x^2 +o(x^2)

ln(1+x) -ln[x+√(1+x^2)] =-(1/2)x^2 +o(x^2) 

分母

 ln(x+√(1+x^2)) = ln(1+x+o(x)) = x+o(x)

ln(1+x) = x+o(x)

ln(1+x) . ln(x+√(1+x^2)) =x^2 +o(x^2)

//

lim(x->0) [ 1/ln(x+√(1+x^2)) - 1/ln(1+x) ]

=lim(x->0) [ ln(1+x) - ln(x+√(1+x^2)) ]/[ ln(x+√(1+x^2)).ln(1+x) ]

=lim(x->0) -(1/2)x^2/x^2

=-1/2

Answer 2

先通分，再利用洛必达法则反复各自求导，直到分子或者分母出现非零的情况。

Answer 3

极限=-1/2，过程如上