Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 13

You need to evaluate the limit, hence, you need to replace 0 for t in the limit, such that:
lim_(t->0) (e^(2t) - 1)/(sin t) = (e^0 - 1)/(sin 0) = (1-1)/0 = 0/0
Since the limit is indeterminate 0/0 , you may use l'Hospital's rule:
lim_(t->0) (e^(2t) - 1)/(sin t) = lim_(t->0) ((e^(2t) - 1)')/((sin t)')
lim_(t->0) ((e^(2t) - 1)')/((sin t)')= lim_(t->0) (2e^(2t))/(cos t)
You need to replace 0 for t, such that:
lim_(t->0) (2e^(2t))/(cos t) = (2e^(0))/(cos 0) = (2*1)/1 = 2
Hence, evaluating the given limit, using l'Hospital's rule, yields lim_(t->0) (e^(2t) - 1)/(sin t) = 2.