<s>[INST] 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.
 [/INST] 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 $0$.
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