Answer
$1$
Work Step by Step
Since we have
$$\lim _{t \rightarrow \infty} \frac{\ln(e^t+1)}{t}=\frac{\infty}{\infty}.$$
We can apply L’Hôpital’s Rule as follows
$$\lim _{t \rightarrow \infty} \frac{\ln(e^t+1)}{t}=\lim _{t \rightarrow \infty} \frac{(e^t/(e^t+1))}{1}=\lim _{t \rightarrow \infty} \frac{e^t}{e^t+1}=\frac{\infty}{\infty}.$$
Again, we can apply L’Hôpital’s Rule as follows
$$ \lim _{t \rightarrow \infty} \frac{e^t}{e^t+1}=\lim _{t \rightarrow \infty} \frac{e^t}{e^t}=1.$$
Hence,
$$\lim _{t \rightarrow \infty} \frac{\ln(e^t+1)}{t}=1.$$
