Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 9

You need to evaluate the limit, hence, you need to replace 0 for x in expression under the limit, such that:
lim_(x->0) (e^(4x) - 1 - 4x)/(x^2) = (e^0 - 1 - 0)/(0) = (1-1)/0 = 0/0
Hence, since the result is indeterminate 0/0 , you may use l'Hospital's theorem, such that:
lim_(x->0) (e^(4x) - 1 - 4x)/(x^2) = lim_(x->0) ((e^(4x) - 1 - 4x)')/((x^2)')
lim_(x->0) ((e^(4x) - 1 - 4x)')/((x^2)') = lim_(x->0) (4e^(4x) - 4)/(2x)
Replacing 0 for x yields:
lim_(x->0) (4e^(4x) - 4)/(2x) = (4e^0 - 4)/(2*0) = (4-4)/0 = 0/0
Hence, since the result is indeterminate 0/0, you may use again l'Hospital's theorem, such that:
lim_(x->0) (4e^(4x) - 4)/(2x) = lim_(x->0) ((4e^(4x) - 4)')/((2x)')
lim_(x->0) ((4e^(4x) - 4)')/((2x)') = lim_(x->0) (16e^(4x))/2
Replacing 0 for x yields:
lim_(x->0) (16e^(4x))/2 = (16e^0))/2 = 16/2 = 8
Hence, evaluating the given limit yields lim_(x->0) (e^(4x) - 1 - 4x)/(x^2) = 8.