<s>[INST] how to find this limit $\lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = -2$ without using L'Hôpital's rule I am looking for simple trigonometric or algebraic manipulation so that this limit can be solved without using L'Hôpital's rule
$$ \lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = -2$$
link on wolframalpha. Thank you for help!!
 [/INST] $$\lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = ?$$
Hint :
$$\cos (x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$$
So: 
$$(\cos x^2 \cos x + \sin x^2 \sin x)=\cos(x^2-x)$$
$$\lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln (\cos(x^2-x))} =  \lim_{x\rightarrow 0} \frac{\dfrac{\sin x^2}{\cos(x^2-x)-1}}{\dfrac{ \ln (1+(\cos(x^2-x)-1))}{\cos(x^2-x)-1}} = ? $$
$$
\lim_{x\rightarrow 0} \dfrac{\sin x^2}{\cos(x^2-x)-1}=\dfrac{\sin x^2}{x^2}\times \dfrac{x^2}{\cos(x^2-x)-1}\times \dfrac{\cos(x^2-x)+1}{\cos(x^2-x)+1} =-2$$
thus :
$$\lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = -2$$
note that :
$$\lim_{x \to 0} \dfrac{\sin x}{x}=1=\lim_{x \to 0}\dfrac{\ln(1+x)}{x}$$
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