I normally leave the calculus to others, but I just love applying L'Hôpital's rule.
L'Hôpital's rule says if we get an indeterminate form like #0/0# or #infty/infty# when we take a limit, we can try again with the derivatives of the numerator and denominator. That is, in the indeterminate case,
#lim_{x to c} f(x)/g(x) = lim_{x to c} {f'(x) }/ {g'(x) } #
Here we have
#lim_{x to 0} {e^x - 1}/sin(2x) quad #
where we initially get the indeterminate form #0/ 0# so we get to apply L'Hôpital. That means we try again using the derivatives of the numerator and the denominator. The derivatives are easy, remembering the derivative of #sin(x)# is #cos(x)#.
#lim_{x to 0} {e^x - 1}/sin(2x) = lim_{x to 0} {e^x}/{2cos(2x)} = {e^0}/{2 cos 0} = 1/2#
