Question

ゆう

176 (1)(3)(5)
177 (1)(3)
178 (1)(3)(5)

分かるとこだけでいいので解説お願いします🙏

182
第6章 数列と極限
問題176 次の極限値を求めよ.
3n
n-3
(2) lim
n→o 2n+1
n→0 4n -+5
n?
(4) lim
n→o 2-n?
n-1
+2
(3) lim
n→0 n2 +1
(5) lim
n→ n3 + n? - 5m +1
4n - 3n - 7
2n3
- 3n + 1
(6) lim
5n3 - n+3
n→0
問題177 次の極限値を求めよ.
7" - 5%
27+2 - 52+1
(2) lim
(3) lim
n→
82
n→0
37 - 57
37 +(-5)
n→0
問題178 次の極限値を求めよ.
5
5
(2) lim
n→0 Vn?-n-n
n→0 Vn? +n-n
5
(3) lim
Vn? - n+n
(4) lim (Vn+1I- Vm)
n→0
n→0
(5) lim (V4n? +n-2n)
Vn+2- Vn
(6) lim
n→0
n→8 vn+1- yn
問題179 次の極限値を求めよ.
COS no
1
2、2
(2) lim
NT
() Sin° nd
sin
(3) lim
n→0
n
/n
n→0
問題 180 次の極限値を求めよ.
3
n→○
(1) lim {log2(4n + 3) - log2(+ 1)}
(2) lim
1+2+…+n
m→0

問題175 (1) 収束(極限値は 3). (2) +0に発散.(3) -00に発散.(4) 振動.
(VR- VR+ 1)
k=1
n
VR- VE+1
=- {(Vi-V2) + (V2-v3) +(V3-V4) ++ (Vn-vn+1)}
=- (VI- Vn+1) = Vn+1- VI= Vn+1-1.
k=1
k=1
基本問題 6.3
(5) 振動、
3
3
3
3n
問題176 (1) lim
lim
n→0 4 +
5
4+0
4°
n→0 4n +5
n
3
-3
1
1-0
1
n
n
lim
(2) lim
n→ 2n -+ 1
1
n→0 2 +
2+0
2°
n
1
0-0
=0.
1+0
n -1
(3) lim
n→○ n +1
n
lim
n
三
1
n→0 1 +
n?
n?
(4) lim
n→0 2
2
+2
1+
lim
1+0
n2
=-1.
ニ
n?
2
0-1
n→0
n2
4n3
3n - 7
4
3
(5) lim
n→o m3 + n?
lim
n→0 1 +
4-0
= 4.
1+0
5n + 1
1
D
5
1
2n°
(6) lim
n→o 5n3
n
n3
3n +1
2
lim
n→0 5
ニ
n +3
3
n3
5°
n2
7"
問題177 (1) lim
5%
n
lim
87
n→0
= 0.
n→0
ニ
1
27+2
(2) lim
3n
57+1
22.2"
lim
5.57
n→0
57
n
5
lim
n→0
37
三
0-5
5%
n
ミ
n→0
ミ5.
0-1
(3) lim
37
n+1
37 +(-5)
37
lim
n→0
n→0
37 +(-5)
0+5
lim
m→0
三
0+1
ミ
問題178 (1) lim
5
Vn? + n
n→0
lim
5(Vn2 + n + n)
三
n
n→0
(Vn? + n
2)(Vn? + n+ n)
II
II
II

問題の解答
5(Vn? + n + n)
lim
(n? + n) - n?
Vn? + n + n
275
=5 lim
n→0
n→0
m
( )-(り
Vn? + n
+1
=5 lim
:5 lim
n2 + n
+1
n→0
n→0
n?
1
1+
=5 lim
=5.2= 10.
n→0
n
5
5(Vn? - n + n)
(Vn2
Vn?
(2) lim
n→0 Vn?
lim
n -
n
m→0
n)(Vn? - n+ n)
5(Vn?
n+ n)
n) - n?
Vn?
= lim
=5 lim
- n +n
(n?
n→0
n→0
-n
=-5 lim
n
n? - n
=-5 lim
n→0
n
n?
n→0
= -5 lim
1
+1
n
= -5.2= -10.
n→0
5(Vn? - n -
n+ n)(Vn? -n-
5
(3) lim
n→ Vn?
= lim
(Vn?
n)
n + n
n→0
5(Vn?
n)
ロ-)
/n?
= lim
=-5 lim
n→0
-n
n→0
n
=-5 lim
= -5.0=D0.
-1
n→0
n
(4) lim (Vn+1- vm) =
(Vn+1-Vm)(Vn+I+Vn)
Vn+I+ Vm
= lim
n→0
n→0
1
= 0.
lim
Vn+I+ Vn
(V4n? +n-2rn) (V4n? +n+2n)
lim
Vn+I+ Vn
n→0
n→0
(5) lim (V4n2+n-2n) = lim
V4n? + n+ 2n
n→0
n→0
n
= lim
= lim
4n2+n+2n
n→0
= lim
V4n2 + n+ 2n
n→0
V4n? + n+ 2n
1
n→0
1
1
lim
1
V4+2
= lim
V4++2
n→0
lim
4nn +2
三
n→0
4n2+n + 2
n→0
11
4
Vn + 2
Vn
Vn + 1
(Vn + 2-V/n) (Vn + 2 + vn)(Vn + I+vm)
= lim
(Vn + I-yn)(Vn +1+ Vn)(Vn+2+ Vn)
(6) lim
n→○