How to calculate: $\lim \limits_{x \to 0}$ $\frac{-1}{\sin^2x}$ + $\frac{1}{x^2}$ How to calculate: $$\lim \limits_{x \to 0}\frac{1}{x^2} - \frac{1}{\sin^2x}  $$
any suggestions what I can do here? 

$$-\frac{1}{\sin(x)^2} + \frac{1}{x^2} = \frac{-x^2+\sin(x)^2}{x^2 \sin(x)^2}.$$
You can use L'Hopital's rule now, but it will take multiple applications since both the numerator and denominator are very small (of order $x^4$, it turns out) near $0$.
A nicer approach is to use the Maclaurin series for $\sin$ to see that 
$$\sin(x)^2=(x-x^3/6+O(x^5))(x-x^3/6+O(x^5))=x^2-x^4/3+O(x^6).$$ 
Consequently
$$-x^2+\sin(x)^2=-x^4/3+O(x^6)$$ 
and 
$$x^2\sin(x)^2=x^4+O(x^6).$$
Now take the ratio and cancel a factor $x^4$ from top and bottom to compute the limit.
