Some calculus, a little better than the usual slopes and distances.
L'Hopital applies because the numerator is sin 0 = 0 and the denominator is ln(2 e^0 - 1) = ln 1 = 0 so we have 0/0, case number one for L'Hopital.
[tex]\dfrac{d\ \sin t}{dt} = \cos t[/tex]
[tex]\dfrac{d\ \ln(2e^t -1 )}{dt} = \dfrac{1}{2e^t - 1} \cdot 2e^t= \dfrac{2e^t}{2e^2 - 1}[/tex]
[tex]\displaystyle \lim_{t \to 0} \dfrac{\sin t}{\ln(2e^t -1)} =\lim_{t \to 0} \dfrac{\cos t }{ 2e^t/ (2e^t - 1) } =\lim_{t \to 0} \dfrac{ (2e^t - 1) \cos t }{ 2e^t}[/tex]
Evaluating the latter form at t=0 we get
[tex]\displaystyle \lim_{t \to 0} \dfrac{\sin t}{\ln(2e^t -1)} = \dfrac{(2e^0 -1) \cos 0}{2e^0} = \dfrac{1}{2}[/tex]
Answer: C 1/2
