Find $\lim_{x \to 0^-}\frac{\sqrt{1 - \cos4x}}{\sin3x - \sin x}$ Find $ \lim_{x \to 0^-}\frac{\sqrt{1 - \cos4x}}{\sin3x - \sin x}$
What I did
$\frac{\sqrt{1 - \cos4x}}{\sin3x - \sin x}$
$\frac{\sqrt{2}\sin2x}{\sin3x - \sin x}$
$\frac{\sqrt{2}\sin2x}{2\cos 2x \sin x}$
$\frac{\sqrt{2}(2 \sin x \cos x)}{2\cos 2x \sin x}$
$\lim_{x \to 0^-}\frac{\sqrt{2}cosx}{\cos 2x} = \sqrt{2}$
But the answer is not $\sqrt{2}$
 A: Square roots are non-negative:  $\sqrt{1-\cos4x}=\sqrt2|\sin 2x|$
So when you take the limit from the left (i.e., with $x\lt0$), you get $\sqrt{1-\cos 4x}=-\sqrt2\sin 2x$.
A: I'm not an expert on limits, Yet graphing this shows that your peaks and asymtopes lie every 1.414 along the x axis, so i would assume root 2 is correct?
A: $$\begin{gathered}
  \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {1 - \cos \left( {4x} \right)} }}
{{\sin \left( {3x} \right) - \sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {2{{\sin }^2}\left( {2x} \right)} }}
{{2\cos \left( {2x} \right)\sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt 2 \left| {\sin \left( {2x} \right)} \right|}}
{{2\cos \left( {2x} \right)\sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - \sqrt 2 \sin \left( {2x} \right)}}
{{2\cos \left( {2x} \right)\sin \left( x \right)}} \hfill \\
   = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 2\sqrt 2 \sin \left( x \right)\cos \left( x \right)}}
{{2\cos \left( {2x} \right)\sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 2\sqrt 2 \cos \left( x \right)}}
{{2\cos \left( {2x} \right)}} =  - \sqrt 2 . \hfill \\ 
\end{gathered} $$
Note that $\sin \left( u \right) > 0,\,\,\forall u \in \left( {0,\frac{\pi }
{2}} \right) \Rightarrow \sin \left( u \right) < 0,\,\,\forall u \in \left( { - \frac{\pi }{2},0} \right)$
