Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 10

You need to evaluate the limit, hence, you need to replace oo for x in expression under the limit, such that:
lim_(x->oo) (e^(4x) - 1 - 4x)/(x^2) = (e^oo- 1 - oo)/(oo) = (oo)/(oo)
Hence, since the result is indeterminate (oo)/(oo) , 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 oo for x yields:
lim_(x->0) (4e^(4x) - 4)/(2x) = (4e^oo - 4)/(2oo) = (oo)/(oo)
Hence, since the result is indeterminate (oo)/(oo) , 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 oo for x yields:
lim_(x->0) (16e^(4x))/2 = oo
Hence, evaluating the given limit yields lim_(x->oo) (e^(4x) - 1 - 4x)/(x^2) = oo.