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.

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}
