Solve the limit using Taylor seris with Big-O notation I have a limit
$$
\lim_{x \to 0} \frac{1}{\sin x} - \frac{1}{x^2}
$$
I've tried to solve it like this:
\begin{align}
&\lim_{x \to 0} \frac{1}{\sin x} - \frac{1}{x^2} =\\
&\lim_{x \to 0} \frac{x^2 - \sin x}{x^2\sin x} = \\
&\lim_{x \to 0} \frac{ x^2 - x + \frac{x^3}{6} - \frac{x^5}{120} - \mathcal{O}\left(x^7\right) }{x^3 - \frac{x^5}{6} + \frac{x^7}{120} + x^2\mathcal{O}(x^7)}
\end{align}

My questions:


*

*How am I supposed to evaluate $x^2\mathcal{O}(x^7)$ in the denominator? I'd say it will be $\mathcal{O}(x^9)$, but I'm not sure about it.

*How am I supposed to evaluate the whole limit with respect to 0 considering Big-O notation? Could you, please, provide some intuitive explanation?


Intuitively. You understand that it is the indeterminate form $\infty-\infty$. To avoid it take $\frac1x$ out of brackets:
$$\frac1{\sin x}-\frac1{x^2}=\frac1x\left(\frac x{\sin x}-\frac1x\right).$$
When $x\to 0^+$, the limit is $(+\infty)(-\infty)=-\infty$. When $x\to 0^-$, the limit is $(-\infty)(+\infty)=-\infty$. Because for $x\to 0$, $x^2$ approaches zero faster than $\sin x\sim x$. 
When you use Taylor expansion:
$$\sin x=x-\frac{x^3}{6}+O\left(x^5\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right),\\
\lim_{x \to 0} \frac{ x^2 - x + \frac{x^3}{6} - \frac{x^5}{120} - O\left(x^7\right) }{x^3 - \frac{x^5}{6} + \frac{x^7}{120} + \underbrace{x^2O(x^7)}_{O(x^9)}}=\\
\lim_{x \to 0} \frac{\require{cancel}\cancel{x}\left( x - 1 + \frac{x^2}{6} - \frac{x^4}{120} - O\left(x^6\right)\right)}{\cancel{x}\left(x^2 - \frac{x^4}{6} + \frac{x^6}{120} + O(x^8)\right)}=-\infty.$$
