Question

求lim[ x^(n+1)-(n+1)x+n]/(x-1)^2 x-->1

Answer 1

①令：x
=
1+t
(t->0)
lim(x->1)
[
x^(n+1)-(n+1)x+n]/(x-1)^2
=lim(t->0)
[
(1+t)^(n+1)-(n+1)(1+t)
+
n]/t^2
【二项式展开，
o(t^2)
表示t^2的高阶无穷小量：（n+1)n(n-1)t^3/3!
+
...
+
t^(n+1)
】
=lim(t->0)
[
[
1
+
(n+1)t
+
(n+1)n/2t^2
+
o(t^2)]
-(n+1)-(n+1)t
+
n]/t^2
=(n+1)n/2
②
罗必塔法则:
lim(x->1)
[
x^(n+1)-(n+1)x+n]/(x-1)^2
=lim(x->1)
[(n+1)x^n
-(n+1)]/2(x-1)
=lim(x->1)
(n+1)nx^(n-1)/2
=n(n+1)/2