Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 12

Given to solve,
lim_(x->-2) (x^2-3x+10)/(x+2)
as x->-2 then the (x^2-3x+10)/(x+2) =0/0 form
so upon applying the L 'Hopital rule we get the solution as follows,
as for the general equation it is as follows
lim_(x->a) f(x)/g(x) is = 0/0 or (+-oo)/(+-oo) then by using the L'Hopital Rule we get the solution with the below form.
lim_(x->a) (f'(x))/(g'(x))

so , now evaluating
lim_(x->-2) (x^2-3x+10)/(x+2)
=lim_(x->-2) ((x^2-3x+10)')/((x+2)')
=lim_(x->-2) (2x-3)/(1)
now plugging the value of x= -2 then we get
= (2(-2)-3)
= -7