Question

Question #a21df

Answer 1

If `x` approaches positive infinity, then the limit is `1`.

If `x` approaches negative infinity, then the limit is `-1`.

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 `x` coefficients on the top and bottom are the same. However, there's a problem with the wording of the question that makes the answer more difficult.

The problem says that the absolute value of `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