Prove that $\lim_{x\to \frac{\pi}{2}}\sin{\frac{x}{2}} \cdot [\sin{x}] = 0$ How can I prove:
$$\lim_{x\to \frac{\pi}{2}}\sin{\frac{x}{2}} \cdot [\sin{x}] = 0$$
I know that for every $0<x<\pi$, where $x\neq \pi/2$,  $[\sin{x}]=0$
How can I go from here?
 A: You said yourself that $[\sin(x)] = 0$ for all $x \in (0, \pi)$ except at $x= \pi/2$.
Then $\sin(x/2)[\sin(x)] = \begin{cases} 0 & x \in (0, \pi) \text{ and } x \neq \pi/2 \\ 1 & x = \pi/2 \end{cases}$.  It follows from the definition of $\lim \limits_{x \to a} f(x) = y$ that the limit as $x \to \frac{\pi}{2}$ of this expression is $0$.  Recall the $\epsilon$-$\delta$ definition:

$\lim \limits_{x \to a} f(x) = y$ means for each $\epsilon > 0$ there is a $\delta > 0$ so that $0 < |x - a| < \delta$ implies $|f(x) - y| < \epsilon$.  Note the $0 < |x -a|$ part, which means we are only considering points near $a$, but not the value $a$ itself, since if $x = a$, then $|x - a| = 0$.

It's your job to now fill in the details/apply this definition based on the hint I gave above.
A: You almost answered the question yourself. The limitand is zero in a neighborhood of the point. Sine is continuous on a neighborhood of the point. What happens at $x=\pi/2$ is irrelevant; it's zero from the right and the left.
A: First of all, defining the domain would be helpful for those of use trying to help you. Are these defined over $\Bbb R$?  Try using the properties of limits: $Lim$(fg)=lim(f)lim(g). Then, we can focus on each limit individually.  Also, as stated by @GFauxPas, analyze the left and right sided limits. Good luck
