Question 436617
To evaluate {{{lim(x->3, (2x^2 - 9x + 2))}}}, note that the function is continuous and defined everywhere. We can substitute 3 to obtain {{{lim(x->3, (2x^2 - 9x + 2)) = 2(3^2) - 9(3) + 2 = -7}}}.


The second one is slightly ambiguous, no indication of whether it is {{{lim(x->2, (2x - (1/3)x + 4))}}} or {{{lim(x->2, (2x - 1/3x + 4))}}}. In either case, both functions are continuous around that interval and you can substitute x = 2.