Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 8

You need to evaluate the limit, hence, you need to replace 0 for x in expression under the limit, such that:
lim_(x->0) (tan 4x)/(x + sin 2x) = (tan 0)/(0 + sin 0) = 0/0
Hence, since the result is indeterminate 0/0, you may use l'Hospital's theorem, such that:
lim_(x->0) (tan 4x)/(x + sin 2x)= lim_(x->0) ((tan 4x)')/((x + sin 2x)')
lim_(x->0) ((tan 4x)')/((x + sin 2x)') = lim_(x->0) (4/(cos^2 4x))/(1 + 2cos 2x)
Replacing 0 for x yields:
lim_(x->0) (4/(cos^2 4x))/(1 + 2cos 2x) = (4/(cos^2 0))/(1 + 2cos 0)
lim_(x->0) (4/(cos^2 4x))/(1 + 2cos 2x) =(4/1)/(1 + 2) = 4/3
Hence, evaluating the given limit yields lim_(x->0) (tan 4x)/(x + sin 2x) = 4/3.