用導數定義求極限
f'(a) = limf(x) = f(a)
x-a
解
解
8.lim √x+1+√√x - I
x-0 √√x+1-√√x+1
x-a
lim
√x+1+
[(√x+1-1) + (√√x-1-1)]/(x-1)
x-0 √√x+1+√√√√x+1 x-0 [√√x+1-1) +(√√x+1-1)]/(x-1)
g(x)=√√x+1, g'(0)=-
h(x)=√√x-1 h'(0) =
59. lim
x-1*
lim
x-1
w√x - wx
√√x - √x
k(x)=√√√x+1, k'(0) =
"√x - √/x_
√√x - √x
3
練習 3.4E
9
r+1
在解特殊型式之極限問題時有其用處。
+
·*. lim √x+1+√√√x −1 2 3
x-0√√√x+1-3√x+1
1 1
2 3
9
x-1
1
m
1
P
1
2₁√√x+1/x = 0
x-1
lim
x-1² √√√x – 1
x-1
1
n
9
1
3/(x-1)² Jx=0
1
3√(x + 1)²
%r+
"x-1_√x-1
=
1
= 2₁
X-
Jx=0
x-1
√x-1
- 1
pq
mn
=
= 5
m
-(n-1)
1
1
3
1
-Xm
m
1
-XP
P
1
1
+ lim
n
Xn
1
-X q
1
√x=1
√x+1= √1-x
3/1-X
!