Answer
$3$
Work Step by Step
Let $\lim\limits_{t \to 0}f(t)=\lim\limits_{t \to 0}\dfrac{\ln [t(1-\cos t)]}{(t-\sin t)}$
But $f(0)=\dfrac{0}{0}$
This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$
This implies that $\lim\limits_{t \to 0}\dfrac{(1-\cos t+t \sin t)}{(1-\cos t)}=\dfrac{0}{0}$
This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$
Thus, $\lim\limits_{t \to 0}\dfrac{(3\cos t-t \sin t)}{\cos t}=\dfrac{3\cos (0)-0}{\cos (0)}=3$
