Prove that $\lim_{x\to {\pi}}\sin{\frac{x}{2}} = 1$

Question

How can I prove $\lim_{x\to {\pi}}\sin{\frac{x}{2}} = 1$ In my book it is said that this can be proved directly from the functions composition limits sentence, Which is defined using those steps:

define $g(x), f(x)$ such that:

1. The following limit exists: $L = \lim_{x\to {x_0}}f(x)$

2. $x_0 = \lim_{t\to {t_0}}g(x)$

3. There is a nominal environment in which $g(t) \neq x_0$

If the above 3 conditions apply, than it is known that:

$\lim_{t\to {t_0}}f(g(t)) = L$

Edit: I understand this can later on be proved easily, yet I'm still not allowed to use the continuous property

Answer 1

We have $\lim_{x \to \pi} x/2 = \pi/2 $. Since $\sin x $ is continuous at $x = \pi $, then

$$ \lim_{x \to \pi } \sin \left( \frac{ x }{2} \right ) = \sin \left( \lim_{x \to \pi } \frac{ x }{2} \right) = \sin \left( \frac{ \pi }{2} \right) = 1 $$

Answer 2

$$sin(\frac{\pi}{2}) = 1 $$