Answer
$0$
Work Step by Step
We have
$$
\lim _{t \rightarrow 0^+}(\sin t)(\ln t)=\lim _{t \rightarrow 0^+}\frac{\ln t}{\csc t}=\frac{\infty}{\infty}
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{t \rightarrow 0^+}\frac{\ln t}{\csc t}= \lim _{t \rightarrow 0^+}\frac{1/ t}{-\csc t\cot t}\\
= \lim _{t \rightarrow 0^+}\frac{-\sin^2t}{t\cos t}=\frac{0}{0}
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$\lim _{t \rightarrow 0^+}\frac{-\sin^2t}{t\cos t}=
\lim _{t \rightarrow 0^+}\frac{-2\sin t\cos t}{\cos t+-t\sin t}=\frac{0}{1+0}=0.
$$
