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This question already has an answer here:

Please help me with this limit without using L'Hôpital's rule. I would by happy if you use simple solving. Thank you as much as I can ;).

$ \lim\limits_{x \to - \infty } {{|\arcsin ({2 \over x})|} \over {\arctan ({5 \over x})}} $

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marked as duplicate by Simon S, Michael Albanese, Marconius, rogerl, Tim Raczkowski Nov 5 '15 at 1:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ This looks remarkably similar to your previous question, and not in a good way. The sooner you change your approach to this site, the better your experience here will be. $\endgroup$ – user147263 Nov 4 '15 at 22:49
  • $\begingroup$ I do not know how to start to solve this limit, because I never seen something like that $\endgroup$ – Marián Slovák Nov 4 '15 at 22:50
  • $\begingroup$ @NormalHuman I had already flagged, asked my officemate to flag. $\endgroup$ – Silvia Ghinassi Nov 4 '15 at 22:55
  • $\begingroup$ @SilviaGhinassi Thanks; it has some close votes now. $\endgroup$ – user147263 Nov 4 '15 at 22:56
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Hint

$$\frac{\arcsin(u)}{\arctan(v)}=\frac{\arcsin(u)}{u}\cdot \frac{u}{v}\cdot \frac{v}{\arctan(v)}.$$ Therefore, if $u(x),v(x)\underset{x\to a}{\longrightarrow} 0$, $$\lim_{x\to a}\frac{\arcsin(u(x))}{\arctan(v(x))}=\lim_{x\to a}\frac{u(x)}{v(x)}.$$

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All you need to know are

(1) $|x| > 0$ if $x \ne 0$.

(2) $\lim_{x \to 0} \frac{f(x)}{x} = 1$ for all the functions and their inverses mentioned in this problem.

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