Page 51 - Phương Trình Mũ Logarit
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Bài toán 4.9: Tìm các giới hạn sau:
             ^     1-cosx.cos2xcos3x  ,  ^
            a)  lim---------^------------------ .ln(jc + e)
                         1-c o sx
            , ^    sin(a + x)sin(a + 2x)-sin^ a
            b)  lim ..................---------------------— .
               “^0          ln(l + 3x)

                                               Giải
                         1-cosx.cos2xcos3x  ,  ,     ^       1-cosx.cos2xcos3x
            a )T a c ó lim ---------^------------------ .ln(x + e)  =lim --------- ^------------------
                     ^-»0      1-c o sx                  '^-^0     1-c o sx
            Vì  1  - cosx.cos2x.cos3x =  1  - cosx + cosx - cosxcos2xcos3x
               =  1  - cosx + cosx(l  - cos2x.cos3x)
               =  1  - cosx + cosx [1  - cos2x + cos2x(l  - cos3x)]
               =  1  - cosx + cosx(l  - cos2x) + cosxcos2x (1  - cos3x)
                                                2        v2
                            -   .   2  kx  í   •  kx^  í    X   ^
                            2  sin  —   sm —         —
                1 -coskx                    2         2
            Và                    2   ...                  .k^
                 1-co sx                  kx         .   X
                            2  sin  -     —         sin —
                                  2    ^   2    ;  l    2 j
                    1 -  cosx.cos2xcos3x
            nên  lim
                 x--> 0    1 -c o sx
                              1 -c o s 2x             1 -cos3x
                 lim 1 + cosx.          + cosx.cos2x.            = 1  +  1.4+ 1.9=  14.
                     V
                 x-> 0         1 -  cos X             1 -c o sx   y
                     1-cosx.cos2xcos3x  ,  ^
            Vây  lim--------—— ------------ ,ln(x + e)  = 14.
                           1  -  cos X

            , ^    sin(ứ + x)sin(a + 2x)-sin^ ữ
            b)  lim--------------------------------------
               xTo          In(l + 3x)
                     sin(í7 + x)sin(ứ + 2x) -  sin^ tìt  ln(l + 3x)
               =  lim--------------------------------------:-------------
                                  X             X
                      sin(a + x)sin(ứ + 2x )-sin " a
            Ta có  lim--------------------------------------
                                    X
                                       2                              1
                                           _
               sin(a + x)sin(a  f 2x) - sin a  = - — [cos(2a + 3x) - cosx] - — (1  - cos2a)
                                                *
                                              1
                                              2                      2
                   1                         1                     3       3
               = - — [cos(2a + 3x) + cos2a] -  — [1  - cosx] = -sin(2a +  — x) sin —  -  sin^ —.
             ^         sin(a + x)sin(a + 2x )-sin ^a
             Do đó  lim----------------^---------------------
                   X->0             X


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