Page 51 - Phương Trình Mũ Logarit

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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