I think I'm supposed to solve this limit using approximations for the behavior when $x$ is close to $0$, but I don't know how to apply the logic for the aproximation.
$$\lim_{x \rightarrow 0} (\frac{1}{sin^2(x)} - \frac{1}{x^2})$$
In the context of this exercise the textbook mentions the following method for approximating: $$f(x) = \frac{1}{1+x+x^2}$$ Since $\lim_{x \rightarrow 0} f(x) = 1$, then $lim_{x \rightarrow 0} [f(x) - 1] =0$; but $f(x)-1 = -x\frac{x+1}{1+x+x^2}$ so, as $x \rightarrow 0$, $f(x) \approx 1-x$. Also for $x$ near $0$ $f(x) - 1 + x = x^3\frac{1}{1+x+x^2}$ so $f(x) \approx 1-x + x^2$.
I think I grasp the logic for it, but I can't apply this to solving limits like the one with $\frac{1}{sin^2(x)}$.
Note: at this point we haven't seen Taylor Polinomials yet, just this method for approximation near $0$.
$$y= \frac{1}{\sin^2(x)} - \frac{1}{x^2}=\frac{x^2-\sin^2(x)}{x^2\,\sin^2(x)}$$ Now, using Taylor series $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\sin^2(x)=x^2-\frac{x^4}{3}+O\left(x^6\right)$$ $$y=\frac{-\frac{x^4}{3}+O\left(x^6\right) } {x^2\left(x^2-\frac{x^4}{3}+O\left(x^6\right) \right) }=\frac{1}{3}+O\left(x^2\right)$$