Evaluate $\lim_\limits{x \to 0}\left[\left(\frac{1}{\sin^{-1}(x)}\right)^2-\frac{1}{x^2}\right],$

Question

$$\lim_{x \to 0}\left[\left(\frac{1}{\sin^{-1}(x)}\right)^2-\frac{1}{x^2}\right]$$ i am a class 12th student and this is my first question on math.SE . This question is part of a previous assignment. I have tried in this way , $$=\lim_{x \to 0}\left[\frac{x^2-(\sin{x}^{-1})^2}{x^2 (\sin{x}^{-1})^2}\right]$$ $$as\; we\; know\; that\qquad \sin^{-1}{x}=\left(x+\frac{x^3}{3!}1^2+\frac{x^5}{5!}1^23^2+\cdots\right)$$ $$=\lim_{x \to 0}\left[\frac{x^2-\left(x+\frac{x^3}{3!}1^2+\frac{x^5}{5!}1^23^2+\cdots\right)^2}{x^2 \left(x+\frac{x^3}{3!}1^2+\frac{x^5}{5!}1^23^2+\cdots\right)^2}\right]$$ $$=\lim_{x \to 0}\left[\frac{1-\left(1+\frac{x^2}{3!}1^2+\frac{x^4}{5!}1^23^2+\cdots\right)^2}{\left(x+\frac{x^3}{3!}1^2+\frac{x^5}{5!}1^23^2+\cdots\right)^2}\right]$$ $$=\lim_{x \to 0}\left[\frac{\left(1-\left(1+\frac{x^2}{3!}1^2+\frac{x^4}{5!}1^23^2+\cdots\right)\right)\left(1+\left(1+\frac{x^2}{3!}1^2+\frac{x^4}{5!}1^23^2+\cdots\right)\right)}{\left(x+\frac{x^3}{3!}1^2+\frac{x^5}{5!}1^23^2+\cdots\right)^2}\right]$$ $$=\lim_{x \to 0}\left[\frac{(-1)\left(\frac{x^2}{3!}1^2+\frac{x^4}{5!}1^23^2+\cdots\right)\left(2+\frac{x^2}{3!}1^2+\frac{x^4}{5!}1^23^2+\cdots\right)}{\left(x+\frac{x^3}{3!}1^2+\frac{x^5}{5!}1^23^2+\cdots\right)^2}\right]$$ $$=\frac{-1}{3}$$ Is this right? and Is there any other way to do it?

Answer 1

First note that $$ \lim_{x\to0}\frac{\sin x}{x}=1. $$ Under $\sin^{-1}x\to x$, one has \begin{eqnarray} &&\lim_\limits{x \to 0}\left[\left(\frac{1}{\sin^{-1}(x)}\right)^2-\frac{1}{x^2}\right]\\ &=&\lim_\limits{x \to 0}\left[\frac{1}{x^2}-\left(\frac{1}{\sin x}\right)^2\right]=\lim_\limits{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}\\ &=&\lim_\limits{x \to 0}\frac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}=\lim_\limits{x \to 0}\frac{\sin x-x}{x^3}\frac{\sin x+x}{x}\frac{x^2}{\sin^2x}\\ &=&2\lim_\limits{x \to 0}\frac{\sin x-x}{x^3} \end{eqnarray} and the rest follows Are all limits solvable without L'Hôpital Rule or Series Expansion.