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so...compute lim as x -->1 for ??? is that [h(x) - h(1)]/(x-1)?

lim x-->1 (x-1) = 0

what happened with 7?

h(x) = x^2-2x+7
h(1) = (1)^2-2(1)+7 = 6
h(x) - h(1) = x^-2x+7-6 = x^2-2x+1

cool