limit $\lim \:_{x\to \:0}\left(\frac{\sin^2\frac{x}{2}}{\sin x}\right)$ how can I solve it?
$\displaystyle\lim_{x\to \:0}\left(\frac{\sin^2\frac{x}{2}}{\sin x}\right)$
 A: Use
$$\sin(x)=2\sin(\frac{x}{2})\cos(\frac{x}{2})$$
So
$$
\lim_{x \to 0} \frac{\sin^2(\frac{x}{2})}{2\sin(\frac{x}{2})\cos(\frac{x}{2})}=\lim_{x \to 0} \frac{\sin(\frac{x}{2})}{2\cos(\frac{x}{2})}=0
$$
A: Just elementary limits:
$$
\lim_{x\to0}\frac{\sin^2(x/2)}{\sin x}=
\lim_{x\to0}\frac{1}{4}\left(\frac{\sin(x/2)}{(x/2)}\right)^{\!2}
\cdot\frac{x}{\sin x}\cdot x
$$
A: because the limit is $\frac{0}{0}$,we can use Lopital Rule
 $$\lim_{x\rightarrow 0}\frac{\sin^2 x/2}{\sin x}=\lim_{x\rightarrow 0}\frac{2(\sin x/2)(\cos x/2)(0.5)}{\cos x}=0$$
A: This problem is a bit more straightforward if we make a simple substitution $t=\frac{1}{2}x$.
$$\lim_{x\to 0}\frac{\sin^2\frac{x}{2}}{\sin x} = \lim_{t\to 0}\frac{\sin^2 t}{\sin 2t} =\dots$$
A: $$\lim_{x\to 0}\left(\frac{\sin^2\left(\frac{x}{2}\right)}{\sin(x)}\right)=$$
$$\lim_{x\to 0}\left(\frac{\frac{\text{d}}{\text{d}x}\sin^2\left(\frac{x}{2}\right)}{\frac{\text{d}}{\text{d}x}\sin(x)}\right)=$$
$$\lim_{x\to 0}\left(\frac{\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{\cos(x)}\right)=$$
$$\lim_{x\to 0}\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\sec(x)\right)=$$
$$\sin\left(\frac{0}{2}\right)\cos\left(\frac{0}{2}\right)\sec(0)=0$$
