Question

How do you find the limit of `abs(6x^2-x-5) / [2x^2+2x-4]` as x approaches 1?

Answer 1

The two sided limit does not exist. For one sided limits see below.

Explanation:

`lim_(xrarr1)abs(6x^2-x-5) / [2x^2+2x-4]` has initial form `0/0`, which is indeterminate.

Since `1` is a zero of the poynomials in the numerator and the denominator, we can be sure that `x-1` is a factor of both.

`abs(6x^2-x-5) / [2x^2+2x-4] = (abs(6x+5)abs(x-1))/(2(x+2)(x-1))`

Observe that

`abs(x-1)/(x-1) = {(1,"if",x > 1),(-1,"if",x < 1):}`

So,

`(abs(6x+5)abs(x-1))/(2(x+2)(x-1)) = {(abs(6x+5)/(2x+4),"if",x > 1),(-abs(6x+5)/(2x+4),"if",x < 1):}`

Therefore,

`lim_(xrarr1^+) abs(6x^2-x-5) / [2x^2+2x-4] = 11/6`

and

`lim_(xrarr1^-) abs(6x^2-x-5) / [2x^2+2x-4] = -11/6`

Because the one-sided limits are distinct, the limit does not exist.