Find $\lim\limits_{x\rightarrow 0} \frac{(\cos{x})^2}{\sin{(x^2)}}-\frac{1}{x^2}$ 
Find 
  $$\lim_{x\rightarrow 0} \frac{(\cos{x})^2}{\sin{(x^2)}}-\frac{1}{x^2}$$

I suppose I have to use L'Hopital's rule here but how?
Edit: I cannot use Taylor yet.
 A: If you can only use L'Hospital, rewrite $$\frac{(\cos{x})^2}{\sin{(x^2)}}-\frac{1}{x^2}=\frac{x^2(\cos{x})^2-\sin{(x^2)}}{x^2\sin{(x^2)}}=\frac uv$$ Now, you will have a funny time since you must apply the rule four times.
I give below the successive derivatives of $v=x^2\sin{(x^2)}$
$$v'=2 x \sin \left(x^2\right)+2 x^3 \cos \left(x^2\right)$$
$$v''=2 \sin \left(x^2\right)+10 x^2 \cos \left(x^2\right)-4 x^4 \sin \left(x^2\right)$$
$$v'''=24 x \cos \left(x^2\right)-8 x^5 \cos \left(x^2\right)-36 x^3 \sin \left(x^2\right)$$
$$v''''=-156 x^2 \sin \left(x^2\right)+\color{red}{24 \cos \left(x^2\right)}+16 x^6 \sin
   \left(x^2\right)-112 x^4 \cos \left(x^2\right)$$ where, for first time, you see a term which is not going to $0$.
A: I suppose it's simpler if you compute 
$$ \lim_{x->0}\frac{x^2(\cos{x})^2 - \sin{x^2}}{x^2\sin{x^2}} $$
Note that you can substitute $x^2$ for $\sin{x^2}$ as x tends to zero and $t$ for $x^2$, the above limit equals to 
$$\lim_{t->0} \frac{t(\cos{\sqrt{t}})^2-\sin{t}}{t^2}$$
Taylor's theorem is indeed convenient, but you can also use L'Hopital's rule twice, which involves some computation. If we are to use L'Hopital's rule, substituting t for $x^2$ reduces loads of work, I suppose. 
The answer, by the way, is -1.
A: Observe
\begin{align}
\frac{(\cos x)^2}{\sin (x^2)} = \frac{(1-\frac{1}{2!}x^2+\frac{1}{4!}x^4+\dots)^2}{x^2-\frac{1}{3!}x^6+\frac{1}{5!}x^{10}-\ldots}=\frac{1-x^2+\ldots}{x^2(1-\frac{1}{3!}x^4+\frac{1}{5!}x^8-\ldots)}
\end{align}
which means 
\begin{align}
\lim_{x\rightarrow 0}\left(\frac{(\cos x)^2}{\sin (x^2)}-\frac{1}{x^2}\right) = -1.
\end{align}
Edit: Rewrite
\begin{align}
\frac{\cos^2x}{\sin(x^2)}-\frac{1}{x^2} = \frac{1-\sin^2 x}{\sin(x^2)}-\frac{1}{x^2}= \left(\frac{1}{\sin(x^2)}-\frac{1}{x^2}\right)-\frac{\sin^2x}{\sin(x^2)}
\end{align}
then observe
\begin{align}
\lim_{x\rightarrow 0}\frac{\sin^2 x}{\sin(x^2)}=\lim_{x\rightarrow 0}\frac{\sin x\cos x}{x\cos(x^2)} = \lim_{x\rightarrow 0}\frac{\sin x}{x}\cdot \frac{\cos x}{\cos(x^2)} = 1. 
\end{align}
I will leave it as an exercise to show that
\begin{align}
\lim_{x\rightarrow 0}\left( \frac{1}{\sin(x^2)}-\frac{1}{x^2}\right)=0.
\end{align}
A: You can simplify your life by noting that 
$$\frac{1}{\sin x}-\frac{1}{x}=x\frac{x}{\sin x}\frac{x-\sin x}{x^3}\to 0$$
So $$\frac{(\cos x)^2}{\sin (x^2)}-\frac{(\cos x)^2}{x^2}\to 0$$
and the limit becomes 
$$\frac{(\cos x)^2}{x^2}-\frac{1}{x^2}$$
 which is much easier, even if you use the hospital rule.
