Solve the limit using Taylor seris with Big-O notation

Question

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}

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My questions:

1. 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.
2. How am I supposed to evaluate the whole limit with respect to 0 considering Big-O notation? Could you, please, provide some intuitive explanation?

Answer 1

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 $\approx -10.7265$ !

Answer 2

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: $$\sin x=x-\frac{x^3}{6}+O\left(x^5\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right),\\ \lim_{x \to 0} \frac{ x^2 - x + \frac{x^3}{6} - \frac{x^5}{120} - O\left(x^7\right) }{x^3 - \frac{x^5}{6} + \frac{x^7}{120} + \underbrace{x^2O(x^7)}_{O(x^9)}}=\\ \lim_{x \to 0} \frac{\require{cancel}\cancel{x}\left( x - 1 + \frac{x^2}{6} - \frac{x^4}{120} - O\left(x^6\right)\right)}{\cancel{x}\left(x^2 - \frac{x^4}{6} + \frac{x^6}{120} + O(x^8)\right)}=-\infty.$$