<s>[INST] Unintentional Negative Sign in Limit Evaluation I've been working on evaluating the following limit:
$$\lim_{x\to 0} \left(\csc(x^2)\cos(x)-\csc(x^2)\cos(3x) \right)$$
According to my calculator, the limit should end up being 4.
Though I've tried using the following process to find the limit, I continue to get -4:
$$\begin{align}
\lim_{x\to 0} \left(\frac{\cos(x)-\cos(3x)}{\sin(x^2)}\right)&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(2x+x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(2x)\cos(x) - \sin(2x)\sin(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \left(1-2\sin^2(x)\right)\cos(x) - 2\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(x)-2\sin^2(x)\cos(x)-2\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(x)-4\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{-4\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\frac{d}{dx}\left(-4\sin^2(x)\cos(x)\right)}{\frac{d}{dx}\left(\sin(x^2)\right)}\right)\\
&=\lim_{x\to 0} \left(\frac{-4\cos^2(x)\cos(x)-\sin^2(x)\sin(x)}{\cos(x^2)}\right)\\
&=\frac{-4\cdot1 - 0}{1} \\
&=-4 \end{align}$$
Is there a specific place in my steps where I am going wrong with respect to negative sign notation or distribution?
(Also, apologies for the extra steps... and possible incorrect ordering of labels, functions, and symbols.)
 [/INST] HINT:
Using Prosthaphaeresis Formulas,
$$\cos x-\cos3x=2\sin 2x\sin x$$
$$\implies \dfrac{\cos x-\cos3x}{\sin (x^2)}=2\dfrac{\sin 2x\sin x}{\sin (x^2)}=2\cdot2\cdot\dfrac{\dfrac{\sin2x}{2x}\cdot\dfrac{\sin x}x}{\dfrac{\sin (x^2)}{x^2}}$$
Now use $\displaystyle\lim_{h\to0}\frac{\sin h}h=1$
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