Calculus: Early Transcendentals, Chapter 2, 2.3, Section 2.3, Problem 13

lim_(x->5)((x^2-5x+6)/(x-5))
sol:
lim_(x->5)((x^2-5x+6)/(x-5))
=>lim_(x->5)((x^2-3x-2x+6)/(x-5))
=>lim_(x->5)(((x-3)(x-2))/(x-5))


In order for the limit to exist , the condition is
lim_(x->5^-)(((x-3)(x-2))/(x-5)) =lim_(x->5^+)(((x-3)(x-2))/(x-5))
let us see whether this condition is satisfied or not.
so, taking
lim_(x->5^-)(((x-3)(x-2))/(x-5))
as x-> 5^- , the denominator x-5 will be a negative quantity approaching to 0
so, lim_(x->5^-)(((x-3)(x-2))/(x-5)) = (((5-3)(5-2))/(0^-))=(6/0^-) = -oo
similarly,
lim_(x->5^+)(((x-3)(x-2))/(x-5))
as x-> 5^+ , the denominator x-5 will be a positive quantity approaching to 0
so, lim_(x->5^+)(((x-3)(x-2))/(x-5)) =(((5-3)(5-2))/(0^+))=(6/0^+)= +oo
as lim_(x->5^-)(((x-3)(x-2))/(x-5)) !=lim_(x->5^+)(((x-3)(x-2))/(x-5))
then the limit does not exist