Question

lim_{t\to0}\frac{\sin t}{\ln(2e^{t}-1)}=
a -1
b 0
c \frac{1}{2}
d 1

Explanation:

Step1: Use equivalent infinitesimals

When \(t\to0\), \(\sin t\sim t\), and \(2e^{t}-1 = 2(1 + t+\frac{t^{2}}{2!}+\cdots)-1=1 + 2t+\cdots\), so \(\ln(2e^{t}-1)\sim2t\) as \(t\to0\).

Step2: Calculate the limit

$$
LATEXBLOCK0
$$

Another way:

Step1: Apply L'Hopital's Rule

Since \(\lim_{t\to0}\sin t = 0\) and \(\lim_{t\to0}\ln(2e^{t}-1)=\ln(2e^{0}-1)=\ln(1) = 0\), it is in the \(\frac{0}{0}\) form.
Differentiate the numerator and denominator: \((\sin t)^\prime=\cos t\), \((\ln(2e^{t}-1))^\prime=\frac{2e^{t}}{2e^{t}-1}\)

Step2: Calculate the limit

$$
LATEXBLOCK1
$$

Answer:

C. \(\frac{1}{2}\)