Answer
$$\ln 2$$
Work Step by Step
Since
\begin{align*}
\lim _{t \rightarrow \infty} \frac{\ln (t+2)}{\log _{2} t}&=\lim _{t \rightarrow \infty} \frac{\ln (t+2)}{\ln t}\ln 2\\
&=\frac{\infty}{\infty}
\end{align*}
then by using L’Hôpital’s Rule
\begin{align*}
\lim _{t \rightarrow \infty} \frac{\ln (t+2)}{\log _{2} t}&=\lim _{t \rightarrow \infty} \frac{\ln (t+2)}{\ln t}\ln 2\\
&=\ln 2\lim _{t \rightarrow \infty} \frac{\frac{1}{t+2}}{\frac{1}{ t}}\\
&= \ln 2\lim _{t \rightarrow \infty} \frac{t}{t+2} \\
&=\ln 2
\end{align*}
