Answer
$2$
Work Step by Step
Let $\lim\limits_{t \to 0}f(t)=\lim\limits_{t \to 0}\dfrac{t-\sin t}{1-\cos t}$
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 $\lim\limits_{t \to 0}\dfrac{(\sin t+t \cos t)}{\sin t}=\dfrac{0}{0}$
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, $\lim\limits_{t \to 0}\dfrac{2(\cos t)-(t) (\sin t)}{\cos t}=\dfrac{2 \cdot 1-(0)(0)}{1}=2$
