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

Awakened
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    Something appears to be missing. Is there supposed to be an argument on the $\sin^{-1}$? Also, do you mean $1/\sin $ or do you mean $\arcsin$? – JoshuaZ Aug 29 '18 at 15:04
  • Please read https://math.meta.stackexchange.com/questions/5020/ – Angina Seng Aug 29 '18 at 15:05
  • Welcome to MSE! We use MathJax to write Math here. Kindly take a quick tour and learn it. Also please add some context and what you have tried so far to the question. Questions with no context are likely to be closed. Thank you. – prog_SAHIL Aug 29 '18 at 15:06
  • Hi and welcome to the site! Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? – 5xum Aug 29 '18 at 15:06
  • See https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion – lab bhattacharjee Aug 29 '18 at 15:10
  • Hint: remember that $\lim_{x \rightarrow 0} \frac{\sin x}{x} =1$. – JoshuaZ Aug 29 '18 at 15:16
  • @JoshuaZ yup I know that, but can anyone solve this please – Awakened Aug 29 '18 at 15:19
  • Ok, next suggestion: Put everything over a common denominator and see what you get. – JoshuaZ Aug 29 '18 at 15:27
  • Please remove the hold , i have formatted the question correctly. @5xum – Awakened Sep 01 '18 at 10:51

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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.

xpaul
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