Download - Lista2 Unificada Limite Continuidade
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7/26/2019 Lista2 Unificada Limite Continuidade
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y = f(x)
limxa
f(x)
limxb
f(x)
limxc
f(x)
x
y
a b c
5
6
3
1
limx3+
f(x) = 5
limx3
f(x) = 5
lim
x2
f(x) = 4
lim
x2
f(x) =
4
limx2
f(x) = 4
f(2) = 4
f
lim
x3+f(x) = lim
x3f(x) = lim
x3f(x)
limxc
(f(x) +g(x))
limxc
f(x)
g(x) =
4; x = 2;; x= 2
limx2
g(x) =g(2) =
f
f(x) =
5; x 17; 1< x 29; x >2
.
limxk
f(x)
k= 1
k= 0.9999
k= 1.0001
k= 2
k= 1.9999
k= 2.0001
cos x
| cos(x)|
|x|
x2
x
log(x)
y= 1 +
x
y= log(x 1) + 2
y = |(x+ 1)(x+ 2)|
limx2
x 2(2
x)(3
x)
;
limx0
x4 +x
x3 + 2x;
lim
x3
x 3x2
4
x
3 x2x2 1 0
x3 1x(x2 4) 0
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p(x) = (x 2)(x+ 3)(1 x)
q(x) = (x 2)2(x+ 1)
r(x) = (3 x)(x 2)2(x 5)
limx0
1
x;
lim
x0
1
x2;
lim
x0
x
|x| ;
limx0
x3
|x| ;
limx2
x2 + 1
x 2;
limx0
x+
1
x
;
limx3+x
x2 9
y= f(x)
f
y = g(x)
g
limx1
q(x) = 0
limx1
3
q(x)=
lim
x1
q(x)
f(x)= 0
lim
x1
q(x)
x2 = 0
g
limx
g(x) = 1
limx
g(x) = 1
limx1+
g(x) = lim
x1g(x) = .
f limx1
f(x) = 2, f(1) = 1
limx1+
f(x) = 2
limx1+
f(x)
limx1+
f(x) =
1 g(x) 2
lim
x3/2g(x)
1
2
limx
cos(
x2 + 1)
x2
f(x) =
9 x2; |x| 3|x| 3; |x| >3. f(x) =
x 1; x 1;
log(x) + 1; x
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lim
x3sen
7
x+ 3
limx2
log |x 2|
limx2
|x 2|(x+ 1)x 2
lim
x5
x+ 3
x+ 5
limx2
|x 2|x2 5x+ 6
limx2
xx2 4 limx1 x3
+ 1x+ 1 lim
x2x+ 2|x| 2 limx1 x
4
2x3
x+ 2x3 + 2x2 x 2 ;
lima2
(a 2)(a2 4)a3 5a2 + 8a 4 limx0
1
x 1
x2
;
lim
x2
x2 3x+ 2x2 3x+ 5
limx1+
x+ 3
1 x limx1x+ 1 2x
x 1 limx1x2 + 2x+ 1
x+ 1
lim
x
2x x23x+ 5
lim
x
3x5 +x 1x5 7 limx
3x3 + 2x4 + 5x5 14x5 3x4 2x2 +x+ 3
lim
x
x2 + 1
x+ 1
limy
5 3y3
8 y+ 10y4
limx
sen
16x6 x+ 1
2x3
x2 + 20
limx0
|x| sen(1/x)
lim
h0
sen(3h)
h
limx
(1 + 1/x)5x
limx/2+
tan(x)
limx0+
(1 2x)1/x
a, b R
c >0
limx0
(1 +ax)b/x
limx
cx2 +ax
cx2 +bx
x 0
h(x) =
1
1 +e1/x
h(x) = 1
x 1|x|
y=
x2 1x 1 y=
1
x2 1 y = x
x2 + 1
y =
x2 1x(x 2) y=
3x2 34 x2
h
x
x3 +x h(x) x
x2 + 1
limx
h(x)
f(x)
|f(x) 3| 2|x 5|4
limx5
f(x)
limx0
(tan(3x))2 + sen(11x2)
x sen(5x) lim
x3x2sen
1
x2
limx0
cos x cos3 x3x2
limh0+
sen(
h) tan(2
h)
5h
limx1
sen
7x+ 1
sen(x/2) 1
(ex1 1)
limh0+
(1 5h3)2/h3
limx
sen x
x limx0sen x
|x|
limxa
f(x)
f
a
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f
a
lim
xaf(x)
f
a
lim
xaf(x) = lim
xa+f(x)
f
A,B,C,D
x
y
A B C D
f(x) =
x; x 0
f(x)> 0
x [0, 1]
g(1)< 0 < g(2)
g
[1, 2]
h
h(2)< k < h(4)
c (2, 4)
h(c) =k
j
k < j(2)< j(4)
c (2, 4)
h(c) =k
f : [3, 1] R
f(3) = 5
f(1) = 2
K [3, 1]
c [2, 5]
f(c) =K
K [3, 4]
c [3, 1]
f(c) =K
K [0, 3] c [3, 1] f(c) =K
a R
R
f(x) =
(x 2)2(x+a)x2 4 x+ 4 ; x = 2
7; x= 2.
f(x) =
2x+ 5 x < 1,
a
x= 1,x2 3
x > 1.
f(x) =
x
|x| ; |x| 1ax;
|x
|0
a; x 0. f(x) =
sen(6x)sen(8x)
; x = 0;a; x= 0.
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a, b R
f
R
f(x) =
ax+b; |x| 2;|x 1|; |x| >2
f(x) =x4 2x3 +x2 + 7 sen(x)
a R
f(a) = 10
b > 0
log(b) =eb
f [0, 1] 0 f(x) 1 c [0, 1] f(c) =c
f
[0, 2]
f(1) = 3
f(x) = 0
x [0, 2]
f(x)< 0
x [0, 2]
f : R R
f(x) Q
x R
f(x)
x R
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3
lim
xbf(x) = 6
limxb+
f(x) = 1
5
c
f(x) =
4; x 3;5; x >3
x 3 4
x 2 4
f(x) =4; x = 2;5; x= 2 x2 4
f(2) = 5 x 3
c = 0 f(x) = sen(1/x) g(x) = sen(1/x)
4
lim
x1f(x) = 5
limx1+
f(x) = 7
limx1
f(x)
5
7
limx2
f(x) = 7 limx2+
f(x) = 9 limx2
f(x)
7
9
1
cos(x)>
0 1 cos(x)< 0 cos(x) =0
x
y
f(x) = cos(x)
| cos(x)|
52
32
2
2
32
52
y= 1
y= 1
x
yf(x) =
|x|
x
x
y
y= 1 +
x
1
log
x
y
y= log(x 1) + 2
1 2
2
1, 2
(x +1)(x + 2)
x
x
y
2 1
y= |(x+ 1)(x+ 2)|
1 1/2
x
0
[3, 1) (1, 3] x31
1
x 1
x3 1 = (x 1)(x2 +x + 1)
a >0
x2 +x+ 10
x 1
x
x2 4
x2 +x + 1
(2, 0) [1, 2)
3, 1, 2
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3 1 2x 2 +x+ 3 + + +1 x + +
0 0 0p(x) + +
x
y
3 1 2
p(x) = (x 2)(x+ 3)(1 x)
1, 2
1 2(x 2)2 + + +
x+ 1 + +0 0
q(x) + +
x
y
1 2
q(x) = (x 2)2(x+ 1)
2, 3, 5
2 3 53 x + +
(x 2)2 + + + +x 5 +
0 0 0r(x) +
x
y
2 3 5
r(x) = (3 x)(x 2)2(x 5)
1
x2 x >0 x2 x
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g
x = 3/2
g(x) = 1
x 3/2 g(x) = 2
1 cos(y) 1
1x2
cos(
x2 + 1)
x2 1
x2.
limx
1x2
= limx
1
x2 = 0,
limx
cos(
x2 + 1)
x2 = 0
x
y
3 3
3
f(x) =
9 x2; |x| 3|x| 3; |x| >3.
x
y
1
1
f(x) =
x 1; x 1;
log(x) + 1; x 2 |x 2| = x 2
x+ 1
x 0 1/x 1/x= 0 x 1
|x|
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x2 +
1 > 0
x
x > 0 x 0 0
x
y
y= x
x2 + 1
x= 0 x= 2
y = 1
x
y
y=
x2 1x(x 2)
x= 2
y = 1
x = 2 x =2
y =3
x
y
y=
3x2 34 x2
x= 2x= 2
y= 31 1
0
x 5 |f(x) 3| 0 f(x) 3
4 3 y =1/x2
1/3
cos
2/5
0
1
e10
y = x
x = + y
sen( + y) =sen cos(y) + sen(y)cos =
sen y
limy0
sen yy
=1
1
x 0+
1 x 0
f(x) =
x
x; x = 0;
2; x= 0;
1
f(0) = 2
f
A
B D
C
C
x = C
2
[2, 3]
[3, 4]
1/5
f(1/2) =10
g
(2, 3)
(3, 4)
K [2, 5]
c [3, 1]
f(c) =K
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K [3, 4]
K [2, 5]
c [3, 1]
f(c) =K
[0, 3] [2, 5]
(x 2)2
a= 5
a = 3 2
a= 1
x= 0
a =
a= 3/4
2a+b = |2 1| = 1,2a+b = | 2 1| = 3. a= 1/2, b= 2 f(0) = 0< 10 lim
xf(x) =
M > 0 f(M) > 10 c [0, M] f(c) = 10
h(x) = log(x)ex
b > 0 h(b) = 0 x 0+ log(x)
ex 1
lim
x0+h(x) =
x
log(x)
ex 0
limx
h(x) = M, N 0< M < N h(M)< 0 h(N)> 0 h d [M, N] h(b) = 0
g(x) = f(x) x
g(c) = 0
f(c) = c g(0) = f(0) 0 = f(0) 0 g(1) = f(1) 1 0
g
c g(1) 0
f(1) = 3 [1, t] f 1 t
c [1, 2]
f(c) = 0
f(x)< 2
[0, 2]
a, b R
a =b f(a) =f(b)
R
R
k
f(a)
f(b)
f
c R f(c) = k
f(x)
x