Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 7

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^(x) - 1)/(tan x) = (e^0 - 1)/(tan 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^(x) - 1)/(tan x) = lim_(x->0) ((e^(x) - 1)')/((tan x)')
lim_(x->0) ((e^(x) - 1)')/((tan x)') = lim_(x->0) (e^x)/(1 + tan^2 x)
Replacing 0 for x yields:
lim_(x->0) (e^x)/(1 + tan^2 x) = (e^0)/(1 + tan^2 0) = 1/(1+0) = 1
Hence, evaluating the given limit yields lim_(x->0) (e^(x) - 1)/(tan x) = 1.