F.5 limits

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圖片參考：http://img339.imageshack.us/img339/8129/88841389.png


2010-11-06 17:47:24 補充：
http://img339.imageshack.us/img339/8129/88841389.png

2010-11-15 21:02:28 補充：
001: Q1: You have read the question incorrectly. One of the term in the denominator is sqrt(4x+1), not sqrt(4x-1)

Q1
lim_(x->2) (sqrt(x+2)-sqrt(3 x-2))/(sqrt(5 x-1)-sqrt(4 x-1))

The limit of a quotient is the quotient of the limits:
= (lim_(x->2) (sqrt(x+2)-sqrt(3 x-2)))/(lim_(x->2) (sqrt(5 x-1)-sqrt(4 x-1)))

The limit of a difference is the difference of the limits:
= (lim_(x->2) sqrt(x+2)-lim_(x->2) sqrt(3 x-2))/(lim_(x->2) (sqrt(5 x-1)-sqrt(4 x-1)))

Using the power law, write lim_(x->2) sqrt(x+2) as sqrt(lim_(x->2) (x+2)):
= (sqrt(lim_(x->2) (x+2))-lim_(x->2) sqrt(3 x-2))/(lim_(x->2) (sqrt(5 x-1)-sqrt(4 x-1)))

The limit of x+2 as x approaches 2 is 4:
= (2-lim_(x->2) sqrt(3 x-2))/(lim_(x->2) (sqrt(5 x-1)-sqrt(4 x-1)))

Using the power law, write lim_(x->2) sqrt(3 x-2) as sqrt(lim_(x->2) (3 x-2)):
= (2-sqrt(lim_(x->2) (3 x-2)))/(lim_(x->2) (sqrt(5 x-1)-sqrt(4 x-1)))

The limit of 3 x-2 as x approaches 2 is 4:
= 0

Q2
lim_(x->0) sin(8 x)^2/x^2

Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(x->0) sin(8 x)^2/x^2 = lim_(x->0) (( dsin(8 x)^2)/( dx))/(( dx^2)/( dx)):
= lim_(x->0) (4 sin(16 x))/x

Factor out constants:
= 4 (lim_(x->0) (sin(16 x))/x)

Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(x->0) (sin(16 x))/x = lim_(x->0) (( dsin(16 x))/( dx))/(( dx)/( dx)):
= 4 (lim_(x->0) 16 cos(16 x))

Factor out constants:
= 64 (lim_(x->0) cos(16 x))

Using the continuity of cos(x) at x = 0 write lim_(x->0) cos(16 x) as cos(lim_(x->0) 16 x):
= 64 cos(lim_(x->0) 16 x)

Factor out constants:
= 64 cos(16 (lim_(x->0) x))

The limit of x as x approaches 0 is 0:
= 64