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!!
 A: Recall the subtraction law 
$$\cos(s-t)=\cos s\cos t+\sin s\sin t.$$
That improves the denominator to $\cos(x-x^2)$. We want to take the logarithm of that. It is nice to express the cosine as $\sqrt{1-\sin^2(x-x^2)}$. Take the logarithm, which is $(1/2)\ln(1-\sin^2(x-x^2))$. So we get a $2$ on top, and want to find the limit of
$$\frac{\sin(x^2)}{\ln(1-\sin^2(x-x^2))}.$$
To finish, you will need to know something about the behaviour of $\frac{\ln(1-u)}{u}$ as $u$ approaches $0$ from the right. If you know that this limit is $-1$, then we can divide top and bottom by $\sin^2(x-x^2)$, and we are nearly finished. I can do the details if the finish is not obvious.
A: 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.
A: It is good recall the following asymptotics.
$$\cos(x^2) = 1 + \mathcal{O}(x^4)$$
$$\cos(x) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4)$$
$$\sin(x^2) = x^2 + \mathcal{O}(x^6)$$
$$\sin(x) = x + \mathcal{O}(x^3)$$
Hence, we get that
$$\cos(x^2) \cos(x) = \left( 1 + \mathcal{O}(x^4) \right) \left( 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4) \right) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4)$$
$$\sin(x^2) \sin(x) = \left(x^2 + \mathcal{O}(x^6) \right) \left( x + \mathcal{O}(x^3) \right) = \mathcal{O}(x^3)$$
Hence, we get that
$$\cos(x^2) \cos(x) + \sin(x^2) \sin(x) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^3)$$
Hence,
$$\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x)) = \ln \left(1 - x^2/2 + \mathcal{O}(x^3) \right)$$
Also, recall that $$\ln(1+t) = t + \mathcal{O}(t^2).$$
Hence, $$\ln \left(1 - x^2/2 + \mathcal{O}(x^3) \right) = -\dfrac{x^2}{2} + \mathcal{O}(x^3)$$
Hence, $$\dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))} = \dfrac{x^2 + \mathcal{O}(x^{6})}{-x^2/2 + \mathcal{O}(x^3)} = \dfrac{-2 + \mathcal{O}(x^4)}{1 + \mathcal{O}(x)}$$
Hence, $$\lim_{x \to 0} \dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))} = \lim_{x \to 0} \dfrac{-2 + \mathcal{O}(x^4)}{1 + \mathcal{O}(x)} = \dfrac{\lim_{x \to 0} \left(-2 + \mathcal{O}(x^4) \right)}{\lim_{x \to 0}  \left(1 + \mathcal{O}(x) \right)} = -2$$
EDIT
Below is a slightly different method.
Note that $$\cos(x^2) \cos(x) + \sin(x^2) \sin(x) = \cos(x^2 - x)$$
We can rewrite $$\dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))}$$ as $$\dfrac{\sin(x^2)}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{x^2}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{x^2}{\ln(1 - 2\sin^2((x^2 - x)/2))}$$
We used the identity that $\cos(\theta) = 1 - 2 \sin^2 \left( \theta/2\right)$.
$$\dfrac{\sin(x^2)}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{-2\sin^2((x^2 - x)/2)}{\ln(1 - 2\sin^2((x^2 - x)/2))} \times \dfrac{x^2}{-2\sin^2((x^2 - x)/2)}$$
Now we have $$\lim_{x \to 0} \dfrac{\sin(x^2)}{x^2} = 1$$
$$\lim_{x \to 0} \dfrac{-2\sin^2((x^2 - x)/2)}{\ln(1 - 2\sin^2((x^2 - x)/2))} = 1$$
$$\lim_{x \to 0}\dfrac{x^2}{-2\sin^2((x^2 - x)/2)} = -2$$
Putting these together again gives us $-2$.
A: $$\lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = ?$$
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 :
$$\lim_{x \to 0} \dfrac{\sin x}{x}=1=\lim_{x \to 0}\dfrac{\ln(1+x)}{x}$$
