How to solve this complex limits at infinity with trig? Please consider this limit question
$$\lim_{x\rightarrow\infty}\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{x\cdot \sin\left(\frac{a}{2x}\right)}$$
How should I solve this? I have no idea where to start please help.
 A: HINT
Note that
$$\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{x\cdot \sin\left(\frac{a}{2x}\right)}=\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{\frac{a}2\cdot \frac{\sin\left(\frac{a}{2x}\right)}{\frac{a}{2x}}}$$
A: Make life easier using $x=\frac 1 y$.
So the problem reduces to 
$$\lim_{y\rightarrow 0}\left(\frac{a\, y}{\sin\left(\frac{a y}{2}\right)} \sin \left(\frac{a}{2}\right) \sin \left(\frac{a}{2}  (y+1)\right) \right)$$ and the first term looks quite familiar.
A: A hint is given in the comment box, another hint is that $\lim_{x\rightarrow\infty}\dfrac{a(x+1)}{2x}=\dfrac{a}{2}$, then $\lim_{x\rightarrow\infty}\sin\left(\dfrac{a(x+1)}{2x}\right)=\sin\left(\lim_{x\rightarrow\infty}\dfrac{a(x+1)}{2x}\right)=\sin\left(\dfrac{a}{2}\right)$.
The fact that one can swipe the limit and the $\sin$ function needs some justification, essentially it is about the continuity of $\sin$ at any point, in this case, it is at the point $a/2$:
$|\sin u-\sin(a/2)|<\epsilon$ for all $|u-a/2|<\delta$, now choose a large $M>0$ such that $\left|\dfrac{a(x+1)}{2x}-\dfrac{a}{2}\right|<\delta$ for all $x\geq M$, then for such an $x$, one has $\left|\sin\left(\dfrac{a(x+1)}{2x}\right)-\sin\left(\dfrac{a}{2}\right)\right|<\epsilon$.
A: Trivial with equivalents: 
as $x\to\infty$,


*

*$\dfrac{a(x+1)}{2x}\to \dfrac a2$, so $\sin\dfrac{a(x+1)}{2x}\sim_\infty \sin \dfrac a2$,

*$\sin\dfrac{a}{2x}\sim_\infty\dfrac{a}{2x} $,
so that
$$\frac{a\sin \dfrac{a(x+1)}{2x}\,\sin\dfrac{a}{2}}{x\cdot \sin \dfrac{a}{2x}}\sim_\infty\frac{a\sin^2\dfrac a2}{x\,\dfrac a{2x}}=2\sin^2\dfrac a2=1-\cos a.$$

A: $$\lim_{x\rightarrow\infty}\frac{a\color{#f00}{\sin(\frac{a(x+1)}{2x})}\cdot \sin(\frac{a}{2})}{\color{#00ff}{x\cdot \sin(\frac{a}{2x})}}\tag{1}$$

$\color{#f00}{\sin\left(\frac{a(x+1)}{2x}\right)} = \sin\left(\frac{a}{2}+\frac{a}{2x}\right);\quad$ so $x \to \infty \implies \sin\left(\frac{a}{2}+0\right) = \sin\left(\frac{a}{2}\right)$ 

L'Hopital: 
$\color{#00ff}{x\sin\left(\frac{a}{2x}\right)} = \left(\frac{\sin \left(\frac{a}{2x}\right)}{\frac{1}{x}}\right) =\left(\frac{-\frac{a\cos \left(\frac{a}{2x}\right)}{2x^2}}{-\frac{1}{x^2}}\right)=\left(\frac{a}{2}\cos \left(\frac{a}{2x}\right)\right);\quad$ so $x \to \infty \implies\frac{a}{2}\cos(0) = \frac{a}{2}$

$$\frac{a\color{#f00}{\sin\left(\frac{a}{2}\right)}\cdot \sin(\frac{a}{2})}{\color{#00ff}{\frac{a}{2}}}\tag{2}$$
$$2\sin ^2\left(\frac{a}{2}\right)\tag{3}$$

$$1-\cos\left(a\right)\tag{3}$$

