How would you work the following limit problems so that the answer for each is (infinity - infinity)?

Question

1. lim [(3/x-2)- (3x/x^2-4)]
x->2+

2. lim [(3/x-2)- (12/x^2-4)]
x->2+

2. lim [(2/x-2)- (6x-x^2x^2-4)]
x->2+

sorry, the third equation is supposed to be
3. lim [(2/x-2)- (6x-x^2/x^2-4)]
x->2+

Answer 1

1)
(3/x-2)- (3x/x^2-4)
= (3/x-2)- (3x/(x-2)(x+2)) = (3/x-2)[1 - x/(x+2)]
now at 2+
= +inf[1-2/4] = +inf[1/2] = +inf

2)
[(3/x-2)- (12/x^2-4)] = [(3/x-2)- (12/(x-2)(x+2))]
=(3(x+2) - 12)/(x-2)(x+2) = (3x +6 -12)/(x-2)(x+2)
=(3x - 6)/(x-2)(x+2) = 3(x-2)/(x-2)(x+2) = 3/(x+2)

at 2+
=3/4


3)

lim [(2/x-2)- (6x-x^2/x^2-4)]

= (2(x+2) - 6x +x^2)/(x-2)(x+2)
=(2x + 4 -6x + x^2)/(x-2)(x+2)
= (x^2 - 4x + 4)/(x-2)(x+2)
= (x-2)^2/(x-2)(x+2)
= (x-2)/(x+2)

at 2+
= 0+/4 = 0+

Answer 2

These all take the indeterminate form infinty-infinity.
We need to change the expression to a fraction getting
lim[6/(x^2-4)] which clearly shows the limit is +infinity.
x --> 2+

For the 2nd problem we get the fraction (3x-6)(x^2-4) so the lim is 0/0. After L'Hospital's rule it becomes 3/2x which again clearly has a limit of +infinity

For the last case the fraction becomes (x-2)/(x+2) which gives a limit of 0.

Answer 3

1=+2

2=lim(3/x-2-12/x^2+4)

=3/2-2-12/4+4=

3/2-1=0.5

3=lim(2/x-2-6x+x^4+4)

=(1-2-12+16+4)=