<s>[INST] 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?

 [/INST] Just as you did
$$y=\frac{1}{\sin x} - \frac{1}{x^2}=\frac{x^2 - \sin x}{x^2\sin x}=\frac{x^2-\left( x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)\right) } {x^2\left(x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right) \right) }$$ that is to say
$$y=\frac{-x+x^2+\frac{x^3}{6}-\frac{x^5}{120}+O\left(x^7\right) } {x^3-\frac{x^5}{6}+\frac{x^7}{120}+O\left(x^9\right) }$$ Now, long division to get
$$y=-\frac{1}{x^2}+\frac{1}{x}+\frac{x}{6}+\frac{7 x^3}{360}+O\left(x^4\right)$$ which shows the limit and how it is approached.
It also gives you a way to estimate the value of $y$ without using the trigonometric functions. Suppose $x=\frac \pi 6$ (which is really far away from $0$). The exact value would be $\approx -1.64756$ while the above expansion would give $\approx-1.64765$. Doing the same with $x=\frac \pi {12}$ and get respectively  $\approx -10.7265$ and $\approx -10.7265$ !
</s>