12.126.
Koshi alomatidan foydalanamiz
0
1
lim
lim
1
lim
1
lim
4
lim
2
2
2
n
arctg
n
n
arctg
n
n
arctg
n
n
n
n
n
n
n
n
n
n
n
n
, ya’ni
0
1
lim
n
arctg
n
hamda
1
ln
2
lim
lim
lim
lim
0
2
ln
2
2
e
x
x
e
x
n
x
x
x
x
x
x
n
n
( Lopital
qoidasiga asosan
0
lim
2
ln
2
lim
x
x
x
x
x
x
).
12.127
. Integral alomatidan foydalanamiz. U holda
1
ln
3
2
1
2
x
x
x
f
funksiyani ko’ramiz. Haqiqatan ham bu
funksiya
;
1
oralig’ida kamayuvchi
va uzluksiz
1
2
1
1
ln
3
2
x
x
dx
dx
x
f
. Bu xosmas integralni tekshirish
uchun solishtirish alomatidan foydalanamiz.
1
ln
1
1
1
ln
3
2
1
2
2
x
x
x
x
. Demak,
2
ln
1
2
ln
1
1
ln
1
lim
1
ln
1
lim
1
ln
1
ln
lim
1
ln
1
1
1
2
1
2
b
x
x
x
d
x
x
dx
b
b
b
b
xosmas integral
yaqinlashuvchi. Shuning uchun berilgan qator ham yaqinlashuvchi bo’ladi.
b)
uzoqlashadi. 12.133.
1
1
x
12.134.
e
x
e
1
12.135.
1
1
x
12.136.
1
1
x
12.137.
1
1
x
12.138.
1
x
ва
1
x
12.139.
1
1
x
12.140.
1
1
x
12.141.
1
x
12.142. har xil
х
12.143.
2
2
x
12.144. har xil
х
12.145.
0
x
12.146
.
0
x
12.147.
0
x
12.151.
1
1
1
2
3
1
,
1
3
n
n
n
n
n
n
x
n
n
u
x
n
n
u
.
1
3
1
2
1
1
1
lim
3
1
1
3
2
3
1
lim
1
3
2
3
1
lim
lim
2
2
1
x
x
n
n
x
n
n
n
n
n
x
n
n
n
n
u
u
n
n
n
n
n
n
n
n
n
.
Demak, qatorning yaqinlashish oralig’i
3
3
x
.
12.152
.
3
3
x
12.153
.
5
5
x
12.154
.
2
3
2
3
x
12.155
. Barcha sonlar o’qida absolyut yaqinlashadi.
12.156
.
1
1
x
12.157
.
3
2
3
2
x
12.158. 1)
0
R
2)
e
R
12.159
.
3
5
x
12.160
.
2
1
x
12.161
.
1
,
1
1
2
x
x
bo’lganda.
12.162
.
1
,
x
arvtgx
bo’lganda.
12.163.
1
,
1
1
2
x
x
x
bo’lganda.
12.164
.
m
x
1
.
12.165
.
2
5
2
5
x
12.166
.
3
3
x
12.167
.
1
,
0
1
,
0
x
12.168
.
1
1
x
12.169
.
3
1
x
12.170
.
0
1
x
12.171
.
1
,
1
1
2
2
2
x
x
x
bo’lganda.
12.172
.
1
1
,
1
ln
x
x
bo’lganda.
12.173
.
1
,
1
2
1
2
x
x
x
bo’lganda
12.174
.
[-1; 1] 12.175.
(-2; 2)
12.176
. [-1; 1)
12.177
. {0}
12.178
. [-
2; 2)
12.179
. (-0,1; 0,1)
12.180.
1
1
x
12.181.
1
x
12.182
. har xil
х
12.183
.
2
2
x
12.184.
har xil
х
12.185.
0
x
12.186.
0
x
12.187.
0
x