Calculus: Early Transcendentals, Chapter 2, 2.3, Section 2.3, Problem 43

lim_(x->0.5^-) (2x + 12)/(|2x^3 - x^2|)
sol:
lim_(x->0.5^-) (2x + 12)/(|2x^3 - x^2|)
=>(lim_(x->0.5^-) (2x + 12))/(lim_(x->0.5^-) (|2x^3 - x^2|)) ------(1)
in the numerator ,we get
(lim_(x->0.5^-) (2x + 12))
= 2(0.5) + 12 = 1 +12 = 13
in the denominator we get
(lim_(x->0.5^-) (|2x^3 - x^2|))
as when x-> 0.5^- so |2x^3 - x^2| is negatiive
so,
|2x^3 - x^2| = -(2x^3 - x^2)= x^2 - 2x^3
so, (lim_(x->0.5^-) (|2x^3 - x^2|)) =(lim_(x->0.5^-) (x^2 - 2x^3))
when approaching to 0 the denominator is a positive quantity so,
(lim_(x->0.5^-) (x^2 - 2x^3)) = 0^+

Now, from (1)
(lim_(x->0.5^-) (2x + 12))/(lim_(x->0.5^-) (|2x^3 - x^2|)) = 13/0^+ = + oo