# Question #a21df

##### 1 Answer

If

If

#### Explanation:

First of all, let's try to factor some things out to make this problem easier.

#2x^2-5x-3 = (2x + 1)(x - 3)#

So our limit becomes:

#lim_(absx->oo)(abs(x-3)(2x-3))/((x-3)(2x+1))#

#(lim_(absx -> oo)abs(x-3)/(x-3))(lim_(absx -> oo) (2x-3)/(2x+1))#

#(lim_(absx -> oo)abs(x-3)/(x-3)) * 1#

The second limit will converge to 1 no matter what, since the

The problem says that the

absolute valueof#x# goes to#oo# .So,

#x# could be approaching either#oo# or#-oo# If

#x# is approaching#-oo# , then#abs(x-3)/(x-3) = -1# .If

#x# is approaching#oo# , then#abs(x-3)/(x-3) = 1# .

So now we have two cases:

CASE 1

#x# approaches#oo#

#lim_(absx -> oo)abs(x-3)/(x-3) = 1# Therefore, the entire limit is 1.

CASE 2

#x# approaches#-oo#

#lim_(absx -> oo)abs(x-3)/(x-3) = -1# Therefore, the entire limit is -1.

*Final Answer*