Limit as $x$ approaches $0$ with constant $a$ Find the limit where $a$ is a constant
$$ \lim_{x \to  0}\frac{\left [ \cos(a+x)-\cos(a-x) \right ]^2}{\tan^2(3x)} $$
I don't know what to do. At first I thought I could replace $a$ with an arbitrary number and then solve the limit but then I got stuck as to how to use the squeeze theorem on this question. Any help?
 A: Hint: $$ \frac{\left[ \cos(a+x)-\cos(a-x) \right]^2}{\tan^2(3x)} = \frac{4\sin^2 a \sin^2 x}{\tan^2(3x)} = \frac{4\sin^2 a }{9}\frac{(3x)^2}{\tan^2(3x)} \frac{\sin^2 x}{x^2}.$$
A: $$
\lim_{x\to0}\frac{\cos(a+x)-\cos a}{x} = \cos'(a) = -\sin a
$$
and
$$
\lim_{x\to0}\frac{\cos a - \cos(a-x)}{x} = \cos' a = -\sin a,
$$
so adding these, we get
$$
\lim_{x\to0}\frac{\cos(a+x)-\cos(a-x)}{x} = 2\cos'a = -2\sin a.
$$
Hence
$$
\lim_{x\to0}\frac{\cos(a+x)-\cos(a-x)}{\tan(3x)} = \lim_{x\to0}\frac{x}{\tan(3x)}\cdot\lim_{x\to0}\frac{\cos(a+x)-\cos(a-x)}{x}.
$$
The first limit in the last expression can be found via L'Hopital's rule or by other methods; the second is what we just found.
Finally, square the whole thing.
A: By the definition of the derivative we have
$$\lim_{x\to0}\frac{\cos(a+x)-\cos(a-x)}{x}=(\cos(a+x)-\cos(a-x))'\big|_{x=0}=-2\sin a$$
and 
$$\lim_{x\to0}\frac{\tan( 3x)}{x}=(\tan(3 x))'\big|_{x=0}=3$$
hence we can see easily that
$$ \lim_{x \to  0}\frac{\left [ \cos(a+x)-\cos(a-x) \right ]^2}{\tan^2(3x)} =\frac{4}{9}\sin^2a$$
A: Even if we only know that $S=\lim\limits_{x\to0}\frac{\sin(x)}{x}$ exists, but don't know it's value (other than $S\ne0$), we have
$$
\begin{align}
\left(\frac{\cos(a+x)-\cos(a-x)}{\tan(3x)}\right)^2
&=\left(\frac{2\sin(a)\sin(x)}{\tan(3x)}\right)^2\\
&=\left(\frac23\sin(a)\cos(3x)\frac{\sin(x)}x\frac{3x}{\sin(3x)}\right)^2\tag{1}
\end{align}
$$
Taking the limit of $(1)$ as $x\to0$ yields
$$
\begin{align}
&\lim_{x\to0}\left(\frac{\cos(a+x)-\cos(a-x)}{\tan(3x)}\right)^2\\
&=\frac49\sin^2(a)\lim_{x\to0}\cos^2(3x)\left(\lim_{x\to0}\frac{\sin(x)}x\right)^2\left(\lim_{x\to0}\frac{\sin(3x)}{3x}\right)^{-2}\\
&=\frac49\sin^2(a)\cdot1\cdot S^2\cdot\frac1{S^2}\\
&=\frac49\sin^2(a)\tag{2}
\end{align}
$$
