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!!

By resorting to some elementary limits we immediately get the final answer. One of them is $\lim_{x\rightarrow 0} \ln(1+x)^\frac{1}{x}=1$ and the other one is $\lim_{x\rightarrow 0} \frac{\cos(x)-1}{x^2}=-\frac{1}{2}.$ Therefore we have that:
$$ \lim_{x\rightarrow 0} \frac{\sin (x^2)}{ \ln ( \cos (x^2) \cos x + \sin (x^2) \sin x)} =\lim_{x\rightarrow 0} \frac{\sin (x^2)}{ \ln ( (\cos (x^2) \cos x)(1 + \tan (x^2) \tan x))}= \lim_{x\rightarrow 0} \frac{\sin(x^2)}{\ln{\cos(x^2)\cos(x)}}=\lim_{x\rightarrow 0} \frac{\sin(x^2)}{\cos(x^2)\cos(x)-1}=\lim_{x\rightarrow 0} \frac{\frac{\sin(x^2)}{x^2}}{\frac{\cos(x)-1}{x^2}+\cos{x}\frac{\cos(x^2)-1}{x^2} }=\frac{1}{-\frac{1}{2}+0}=-2.$$ 
REMARK: 
 $$\lim_{x\rightarrow 0} \frac{\cos(x)-1}{x^2}=\lim_{x\rightarrow 0} \frac{\cos(x)-1}{x^2} \frac{\cos(x)+1}{\cos(x)+1} = \lim_{x\rightarrow 0} \frac{-\sin^2(x)}{2x^2}=-\frac{1}{2}$$ (no use of L'Hopital Rule)
Also notice that:
$$\lim_{x\rightarrow 0} \frac{\sin(x^2)}{\ln{\cos(x^2)\cos(x)}}=\lim_{x\rightarrow 0} \frac{\sin(x^2)}{\frac{\ln(1+(\cos(x^2)\cos(x)-1))}{\cos(x^2)\cos(x)-1} (\cos(x^2)\cos(x)-1)}=\lim_{x\rightarrow 0} \frac{\sin(x^2)}{ \cos(x^2)\cos(x)-1}$$
Here we apply the trivial limit: $\lim_{u\rightarrow 0} \frac{\ln(1+u)}{u}=1$
The proof is complete.
