Limits going wild

$\displaystyle\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2(x)}\right)=\lim_{x\to 0}\frac{\sin^2(x)-x^2}{x^2\sin^2(x)}$ $=\displaystyle\lim_{x\to 0}\frac{\frac{x^1\cdot x^1}{1!1!}-2\frac{x^1\cdot x^3}{1!6!}+x^2}{x^2\frac{x^1\cdot x^1}{1!1!}}=\lim_{x\to 0}\frac{-\frac{1}{3}x^4}{x^4}=-\frac{1}{3},$

so the answer is $1+3=\boxed{4}$ .

$L = \displaystyle \lim_{x \to 0} \left(\dfrac{1}{x^2} - \dfrac{1}{\sin^2x} \right)$

$\Rightarrow L = \displaystyle \lim_{x \to 0} \left(\dfrac{1}{x^2} - \dfrac{1}{\left(x - \frac{x^3}{3!}\right)^2} \right)$

$\Rightarrow L = \displaystyle \lim_{x \to 0} \dfrac{1}{x^2}\left(1 - \dfrac{1}{\left(1 - \frac{x^2}{3!} \right)^2} \right)$

$\Rightarrow L = \displaystyle \lim_{x \to 0} \dfrac{1}{x^2} \left(1 - \left(1 - \dfrac{x^2}{3!}\right)^{-2} \right)$

$\Rightarrow L = \displaystyle \lim_{x \to 0} \dfrac{1}{x^2} \left(1 - \left(1 + (2)\dfrac{x^2}{6} \right) \right)$

$\Rightarrow L = \displaystyle \lim_{x \to 0} \dfrac{1}{x^2} \left(-\dfrac{x^2}{3} \right) = \boxed{-\dfrac{1}{3}}$