Answer
$f'(\pi)$ where $f(t)=\sin t\cos t$
Work Step by Step
The derivative of $f(t)$ at a point $a$ is defined as
$f'(a)=\lim\limits_{t \to a}\frac{f(t)-f(a)}{t-a}$
Comparing $\lim\limits_{t\to \pi}\frac{\sin t\cos t}{t-\pi}$ with the above definition, we get
$a=\pi$
$f(t)-f(a)=f(t)=\sin t\cos t$ and
$f(a)= f(\pi)= \sin\pi \cos \pi=0$
Therefore, the given limit can be expressed as the derivative of $f(t)= \sin t\cos t$ at the point $ t= \pi$.
