## Calculus (3rd Edition)

$f'(\pi)$ where $f(t)=\sin t\cos t$
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 $t= \pi$.