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