-
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m=0
画数(W)兩E Ill
x-a
Aulim
ho
enzo
ho
x-a
h
xr
一
3m
h
ho zh
xto
x gix) 1x/
lim
hoxo-h(c)
Xtc
-a
Podcast
X-
x-c
you
XCL
X-6
x-
h
(D)若函數 (x)滿足 lim
xg(x)
-=1, al lim g(x) TFT
lim feaths fraz, lim tial fan lim feath) fial, limtial frach
zm F
hyo Cathi-a
-
7X/
R 15 ) : gux?
limitel, he lim 1 .
(C)若函數 (x)滿足 in
in h(x) – h(C) **, ni lim h(x) = h(c) ()
17 lim (X-4 to, solim [hox - hcc] -
x-c
(B)若 L(x) 與 (x)均為實數對應到實數的函數,則ko/(x)=lok(x)
G
hexy-h(6)
lim hex)=
(X-() = lim axo =0=7
k(x)
(
XC hlu)
/ (E)若函數 (x)為實係數多項式函數且 lim
=5,則函數 (x)的次數必
(x − 3)(x - 2)
超過一次,
3. 下列哪些函數在x=0可以微分? (A)f(x) = px (B)g(x) = xsinx
sing
lim Ex-o.1]-(1)
1
©h(x) = xsin DJK(0)T– 0.7 É) 16x) = {2x+1, x20
[x² – 3, x<0 in
singan
4.下列關於無窮數列與級數的收斂敘述,請選出正確的選項: (A)* * * 51(an) He tot, al lima, =0 (B)
12.
glys-geol xring- o
o
1
asino
X
( EES
n>
. 以利别的英文單字。
...
n-00
-00
n=1
n=1
n->00
1-00
7-00
若 lim a = 0,則無窮級數n = 0 (C)若無窮級數, 存在,則im can = 0 (D)設《an》、《bo》為兩個無窮
7.1.-3...
數列,則lim(a + b) = lim an + limb, 恆成立 (E)設〈an + ba〉、an-bo》均為收斂數列,則{an》、《bb》亦均為收
斂數列.
neho
8 9 4 3 2 1 1 os, & ant t3 th & Fan I takto
二、填充題:每格5分,共65 分,/hp
1. 計算下列無窮級數的和:
n=no
00
neho
nal
s
1
2
3 7 2" -1
+
+...+
+...=
25 125 5"
2
子
Śn
以上
(2) 3 (2n-1)(2n+1)
|
$
1
-
to ,
X4 x Xtroy());
1.9
si
zht/
2.
enfr
4 ontrout 4 enp19
4 4-7 yote
nchillanti)
6
求下列極限值:
any 30th
ninti) (2011)
6langany)
- (nn) Lanti) = 2n + 3n²th
211721 ..th) =
ſnith.
nx intl)
1² +2²+...+n²
(1) lim
2n²+2n-1
2 horny
lim(V2+4 +6+ ... +2n - V1+3+...+(2n+1)) =
n-
n-00
