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

1 Answer
Apr 25, 2016

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.