Find $\lim_{x \to 0^-}\frac{\sqrt{1 - \cos4x}}{\sin3x - \sin x}$

Question

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 $\sqrt{2}$

Answer 1

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 $\sqrt{1-\cos 4x}=-\sqrt2\sin 2x$.

Answer 2

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?

Answer 3

$$\begin{gathered} \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {1 - \cos \left( {4x} \right)} }} {{\sin \left( {3x} \right) - \sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {2{{\sin }^2}\left( {2x} \right)} }} {{2\cos \left( {2x} \right)\sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt 2 \left| {\sin \left( {2x} \right)} \right|}} {{2\cos \left( {2x} \right)\sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - \sqrt 2 \sin \left( {2x} \right)}} {{2\cos \left( {2x} \right)\sin \left( x \right)}} \hfill \\ = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 2\sqrt 2 \sin \left( x \right)\cos \left( x \right)}} {{2\cos \left( {2x} \right)\sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 2\sqrt 2 \cos \left( x \right)}} {{2\cos \left( {2x} \right)}} = - \sqrt 2 . \hfill \\ \end{gathered} $$ Note that $\sin \left( u \right) > 0,\,\,\forall u \in \left( {0,\frac{\pi } {2}} \right) \Rightarrow \sin \left( u \right) < 0,\,\,\forall u \in \left( { - \frac{\pi }{2},0} \right)$