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329.Сборник задач для самостоятельного решения по теме Предел функции

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y
y
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y
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−3,5
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y = f ( 12 x)
y = kx
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y = 2x + 3
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y = f (x)
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y = − x2 − 1
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x
y = 21 (x + |x|)
y = sin x + | sin x|
y = | lg x|
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y = y0 + a(x − x0 )
y = 21 x2 + x + 1
y = x2 − 3x + 2
y = x3 − 3x + 2
y = x2 − x4
1−x
1+x
3x+2
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y =3−
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n
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n
√
3
1
n
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1
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n=1
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p
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n
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2
3
n
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n
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f (x) = arcctg x
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x
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f (x) − f (x0 )
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1+cos πx
2
x→1 (x−1)
ln cos 2x
lim
x2
x→0 sin
ex −e
lim cos
πx
2
x→1
1
lim (2x − 1) x−1
x→1
lim
"
"
B
lim (x2 − x − 2) = ( lim x)( lim x) − lim x − lim 2 = −2.
x→1
sin 5x−sin 3x
sin x
cos x
lim 1−2
x→π/3 π−3x
lim
3
*
;S
%
P
3
3
K
K
H
√
3
x→0
2
lim 2x −4
x→1 x −x−2
" R "
B
/ / / /
/ + + +
,' ,- ,> ,
/
*
,7
/
*
,+
=
x→x0
V
ln cos x
x→0 x arcsin x
lim
lim (f (x) ± g(x))
x→x0
100
lim x 50 −2x+1
x→1 x −2x+1
B B
x→0
/ /
,7 ,0
lim x2 − 3
x
x1
x
lim 2 +3
2
x→2
B
1
2−x
/
,
/ / /
,* ,+ ,7
B
B
B B
ln x
πx
x→1 cos 2
e−x
lim
x→e 1−ln x
x
x
lim 2 −3
x→0 sin x
lim
/
*
,0
sin x−1
cos x
/ /
* *
, ,*
B
B
B
B
x→+∞
√
x+3−2
lim √2x+7−3
x→1 √
1−x−3
√
lim 2+
3
x
x→−8
√
√
1+sin x− 1−sin x
lim
x
x→0
lim x sin x1
x→0
ln(1+3x )
x
x→−∞ ln(1+2 )
lim 2ctg x
x→−0
lim arcctg x1
x→−0 x
sin2 x
lim
2
x→+∞
3
lim x5 −2x−1
x→−1 x −2x−1
lim
B
p
√
√
x+ x+ x− x
lim arcsin 1−x
1+x
x→−∞
B B
B
B
lim x5 −3x+2
x→1 x −4x+3
n
n−1
+...+x−n
lim x +x x−1
x→1
q
/ / /
/ / /
,> , ,/
B
B
B
B
4
x→ 2
"
!
ln(1+3x )
x
x→+∞ ln(1+2 )
lim 2ctg x
x→+0
lim arcctg x1
x→+0
√
lim 1−x
x .
x→1
lim
B B
/ / / /
/ + + +
,7 ,0 , ,*
/ /
/ /
,* ,+
/
/
,
/
/
,0
/
/
,/
*
,'
/
*
,/
/ /
* *
,> ,
/ / /
*
,/ ,' ,/
,
/ /
,- ,>
lim arcsin 1−x
1+x
limπ
lim f (x) = a
x→x0
x→x0
x→+∞
lim
%T
f (x)
;
?
*
()
b+/
Copyright ОАО «ЦКБ «БИБКОМ» & ООО «Aгентство Kнига-Cервис»
x→2
lim
lim (x + 2)
x2 − 4
x+2
4
x→2
= lim
=
= .N
− x − 2 x→2 x + 1
lim (x + 1)
3
x→2 x2
x→2
x→0
.
.
B
lim xx
x
lim 1 + x2 .
x→∞
x+2
lim x−1
.
x+3
x→∞
1
lim (cos x) x ;
x→0
+
,0
/(*
*C
=;
$;
(
=1
W
R
ln(1+x)
x
B
lg(1+10x)
x
q
1
ln 1+x
1−x
x→0 x
ln(x−1)
lim
x→2 sin πx
lim
B
/ /
' '
,> ,
B
x→+∞
x)
lim ln(cos
x2
x→0
lim
x→0 B
B
lim x[ln(x + 1) − ln x]
B
/ /
' '
, ,*
lim [ln(2x + 1) − ln(x + 2)]
x→+∞
/
'
,/
;
?
*
()
b+/
1
1
ln(1 + x)
= lim ln(1 + x) x = ln lim (1 + x) x = ln e = 1.N
x→0
x→0
x
B
x→0
t→0
V
3
lim
e −1
t
= lim
.
t→0 ln(1 + t)
x
B
B
B
eax −ebx
x
sh x
lim
x→0 x
1
lim (ex + x) sin x
x→0
lim
x→0
B
B
> 0)
B
∗
ax −1
x (a
1−e−x
lim
x→0 sin x
lim chxx−1
2
x→0
lim
x→0
/ / /
+ + '
,/ ,' ,-
; 3
?"
* () b+/ 3
2
x→0
/ / /
+ + '
+ 7 0
, , ,
√
x− a
x−a
1
1
√
√ = lim √
√ = √ .N
= lim
x→a (x − a)( x +
x→a
x−a
a)
x+ a
2 a
=1
x
3
#
2
1
P
a>0
ex −1
x
t = e − 1 x = ln(t + 1)
"
"
"
H
x
x→0
P
/(*
*C
=;
$;
(
K
KG
lim
Y
B
+
,0
+
,7
√
x− a
x−a
A:8
√
O$
%&
'%
=;
K
K
x→a
√
lim
x→0
3 "
"
" 3
3
"
1 3
" 2 lim
lim
/
'
,+
R
B
B
√
x−8
√
3
x→64√ x−4 √
3 2
−2 3 x+1
lim x(x−1)
2
x→1
lim
"
lim
.
sinx x
lim (cos x) x2 .
#
"
P
K
K
" *S
%;
;
?
*
()
b+/
+
,>
+
,-
x−1
x−1
x→1 √
3 x−1
lim √
4
x→1 x−1
x→a
x+1
x2 −2x+3
x2 −3x+2
x x
lim 1+2
2
x→−∞
lim
x→+0
1
x−1
x2 −1
x→0
V
5
3
" 1
3
2 " "
√
1+x−1
√
3 1+x−1
√
H
1
lim (1 + sin x) x .
x→0
/ /
' 7
,7 ,0
x→∞
x→1
/ /
7 7
,+ ,7
lim x .
2 x2
x +2
lim 2x
.
2 +1
x→∞
1 x
lim x + x
x→+0
x
x
lim x+1
.
x→+0
x→0
1+x = y
√
1+x−1
y2 + y + 1
3
y3 − 1
lim √
= lim
= .N
= lim 2
3
x→0
y→1
y→1 y − 1
y+1
2
1+x−1
lim
.
lim
H
"
lim
x→0
6
2x
K
KG
y→a
1 x+1
x2
ln x
.
P
x0
lim
x→∞
x
/ /
7 7
, ,*
x→+∞
2x−1
3x+2
B
/ +
7 ,' ,-
B
/ /
7 7
, ,/
lim f (ϕ(x)) = lim f (y) = b.
x→x0
H
%X
;
?
*
()
b+/
+
,>
B
1
x2 −3x+2
lim
B
B
B
+
lim f (y) = b,
"
2
2x−x2
y→a
f (ϕ(x))
"
x→2
lim ϕ(x) = a
x→x0
lim
/ / /
' 7 7
,' ,- ,>
3
1−x3
−
2
1
K
I
K
B B B
B B
+ + + + +*
, ,* ,+ ,7 ,0
x→1
1
1−x
x3 +1
2
x→−1 x +1
2
lim 2x −2x
x→2 x −4x+4
3
lim xx4 −3x+2
−4x+3
x→1
(a+x)3 −a3
lim
x
x→0
lim
lim
lim
%%
+ + + +
* *
,/ ,' ,- ,>
x2 −4
2 −3x+2
x
x→2
2
lim 2x −1
x→−1 x +3x+2
2
lim x x−5x+10
2 −25
x→5
x2 −(a+1)x+a
lim
x3 −a3
x→a
+
,
;
?
*
()
b+/
Copyright ОАО «ЦКБ «БИБКОМ» & ООО «Aгентство Kнига-Cервис»
N
=e
lim
1
ln(1+x)
x
%M
;
?
*
()
b+/
B
B
B
6
x+1
2x + 1
x2
" lim
x→0
H
lim
#
1R
cos 3x − cos 7x
= 20.N
x2
arctg x
x
y = arctg x
lim
x→0
x→0
= 0.N
B
+ +
> >
,+ ,7
sin 2x
= 2,
x
lim
x→0
lim x2 = +∞.
B
B
B
sin 5x
= 5,
x
1
=
2
1
x
1
x
x→∞
lim
B
+ + + +
,0 , ,* ,+
B
B
V
lim
x→0
1+
x+1
lim
= lim
x→∞ 2x + 1
x→∞ 2 +
x→∞
cos 3x−cos 7x
x2
sin 5x sin 2x
cos 3x − cos 7x
=2
x2
x
x
Y
= 21 = 2.N
t
"
x+1
2x+1
1+x
lim
x→0
*
%;S
P
sin 2x
x
x2
K
K
x→∞
H
t = 5x
sin t
sin 5x
sin t
= lim t = lim
· 5 = 1 · 5 = 5.N
t→0
t→0
x
t
5
lim
sin 5x
x
3
*S
%;
x→0
P
K
KG
G
x→0
W
R
lim
= e0 = 1.N
lim (1 + x) = 1
x→0
lim
x→0
sin x
= 1.
x
P
x→0
lim ( sinx2x ) = 2
lim
H
H
K
K
sin 2x 1+x
x
lim
*S
%;
K
KG
x→+∞
lim
x→+∞
P
=0
3
R
lim 1
x→+∞ x
21
lim (1 + x) = +∞
x→+∞
x→0
x→0
lim (1 + x) x
x→+∞
lim (1 + x) x = ex→+∞
H
lim
1
= 2 .N
e
P
*S
%;
K
KG
H
1
x
lim −2x
= ex→∞ x+1 = e−2 .
K
K
lim
x→∞
x−1
x+1
x→∞
x→+∞
p
x(x + a) − x
√
x2 − 5x + 6 − x
"
2
lim
3 " 4
3 !
"
"
− x+1 ! −2x
x+1
2
1−
x+1
lim
x→∞
√
√ x+a− x
x→+∞
√
lim x x2 + 1 − x
x→+∞
√
lim x + 3 1 − x3
lim
"
2
lim
x→0
x−8
√
3 x−2
√
√5+x
lim 3−
5−x
x→4 1−
√
√
2
x2 +2x−6
lim x −2x+6−
2 −4x+3
x
x→3 √
√
3
3 a
lim a+x−
x→0
x
lim
x→8
− x+1
lim
x→0
B B
2
x+1
x→∞
x→∞
1−
.
lim
x→1
*S
%;
lim
2
x→7
√
2− x−3
x2 −49
√
x−1
√
3
x−1
√
√
1+x− 1−x
x
√
√
x+a− a
x
B
"
x→∞
2
x+1
1−
− x+1 ! −2x
x+1
lim
*
%;S
P
lim
"
2
x+1
V
=
=
+ + + +
,> , ,/ ,'
lim
x→∞
2
1−
x+1
x
+ + +
> > ,/ ,' ,-
=
x−1
x+1
x
1∞
lim
x→∞
%R
Y
3 3 R "
Copyright ОАО «ЦКБ «БИБКОМ» & ООО «Aгентство Kнига-Cервис»
arctg x
y
cos y
= lim
= lim sin y .
y→0 tg y
y→0
x
y
x→0
y→0
sin 2y cos y
·
.
sin y cos 2y
x→x0
M
B B
x→x0
lim α(x)g(x)
=
ex→x0
=
x→x0
e
=
lim (f (x)−1)g(x)
α(x)g(x)
=
.
lim f (x) = +∞
x→x0
lim g(x) = 0,
x→x0
3
R
21
B
−x
1
B
lim (g(x) ln f (x))
lim (f (x))g(x) = lim eg(x) ln f (x) = ex→x0
x→x0
R
lim
x→∞
4
"
1
x
P
x−1
x+1
"
x→∞
H
lim (g(x) ln f (x))
x→x0
lim
.
x→x0
5 W
" "
tg x−sin x
x3
arctg 2x
lim
x→0 sin 3x
2
lim 1−xπx
x→1 sin √
x
lim 1− xcos
2
x→0
lim
x→0
1
(1 + α(x)) α(x)
B B
B
x→0
π
2
lim f (x) = 0
x→+∞
√
√
sin x + 1 − sin x − 1
lim ctg 2x ctg
lim
x→x0
K
K
*S
%;
"
lim
lim n sin πn
n→∞
x−sin a
lim sin x−a
x→a
1−sin x
lim π−x 2
x→π
lim sin x−cos x
x→π/4 1−tg x
T
lim
x→1
sin 3x
x
lim sinx x
x→∞
lim
x→0
B B
B
B
lim
x→0
=
x→x0
B
lim
x→0
B B
x→0
B B
lim
B
x→1
f (x) = 1 + α(x)
lim (f (x))g(x)
+ + + + + + + + + + +
- - - 0 0 0 0 0 > > >
, ,/ ,' ,- ,> , ,/ ,' ,- ,> ,
B
B
B B B
B
πx
2
cos x−cos 3x
x2
arcsin x
x
x−sin 2x
x+sin 3x
cos πx
√2
1− x
lim (1 − x) tg
f (x)
lim α(x) = 0
;
?
*
()
b+/
+
>
,*
+ + + +
0 0 > >
,+ ,7 ,0 ,
+ + + + + +
- - - 0 0 0
,* ,+ ,7 ,0 , ,*
+
,
∗
sin x
x
5x
lim sin
x→0 sin 2x
lim sin πx
x→1 sin 3πx
x
lim 1−cos
x2
x→0
a
lim cos x−cos
x−a
x→a
lim tg πx
x→−2 x+2
lim
lim g(x) = ∞,
x→x0
3
"
1" 3
4
3
=1
sin 2y cos y
sin 2y
cos y
lim
·
= lim
· lim
= 2.N
y→0 sin y
y→0 sin y
y→0 cos 2y
cos 2y
x→2
lim f (x) = 1
x→x0
− 2y ctg y = lim
2
= e,
x→x0
lim cos y
y→0 cos 2y
y→0
=2
V
lim sin 2y
y→0 sin y
− x = lim ctg
π
x
R
y = π/4 − x
1
x
lim (f (x))g(x)
0·∞
"
ctg(π/4 − x).
1+
5 K
" " K
" *
S
" %;
"
" "
"
4 1 3
1 P
ctg 2x
x→∞
e = 2,718281828459 . . .
x → π/4
H
4
%[
1
lim (1 + x) x = lim
lim ctg 2x · ctg(π/4 − x)
lim ctg 2x · ctg
= 1.N
sin y
y→0 y
lim
y→0
x→π/4
x→π/4
lim cos y
=
y→0 sin y
y
cos y
"
lim
π
=1
sin y
y
3
3 " " " 3 "
%O
lim
y→0
TDE
lim cos y = 1
y→0
V
Copyright ОАО «ЦКБ «БИБКОМ» & ООО «Aгентство Kнига-Cервис»
1−
x−1
= lim
x→∞
x+1
1+
1
x
1
x
= 1.
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